Q.Using binomial theorem, prove that $6^n - 5n$ always leaves remainder 1 when divided by 25.

Showing the 12 most recent of 30 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If $\alpha_n$ is the coefficient of $x^n$ in the expansion of $(1-x)^{-5}$ and $\beta_n$ is the coefficient of $x^n$ in the expansion of $(1-x)^{-4}$, then $\alpha_{12} + \beta_{13} =$ (A) $\alpha_{13}$ (B) $\beta_{13}$ (C) $\alpha_{25}$ (D) $\beta_{25}$

   ›Reveal solutionSolution

   The coefficients come from the binomial series for negative exponents: $\alpha_n = \binom{n+4}{4}$ and $\beta_n = \binom{n+3}{3}$.

   We compute $\alpha_{12} + \beta_{13} = \binom{16}{4} + \binom{16}{3} = \binom{17}{4} = \alpha_{13}$, so the answer is (A).

   ---

   Concept and Intuition

   The expansions of $(1-x)^{-k}$ for positive integer $k$ are given by the negative binomial series:

   $(1-x)^{-k} = \sum_{n=0}^\infty \binom{n+k-1}{k-1} x^n$

   Here, $\binom{n+k-1}{k-1}$ counts the number of ways to write $n$ as a sum of $k$ nonnegative integers — that’s why it appears.

   So $\alpha_n$ for $(1-x)^{-5}$ is $\binom{n+4}{4}$, and $\beta_n$ for $(1-x)^{-4}$ is $\binom{n+3}{3}$.

   The problem asks for $\alpha_{12} + \beta_{13}$. The trick is to notice that adding two consecutive binomial coefficients of the same upper index gives the next binomial coefficient — Pascal’s rule. That will let us rewrite the sum as a single coefficient from the original series.

   ---

   Step-by-step

   1. Write the coefficients explicitly For $(1-x)^{-5}$:

   $\alpha_n = \binom{n+5-1}{5-1} = \binom{n+4}{4}$

   For $(1-x)^{-4}$:

   $\beta_n = \binom{n+4-1}{4-1} = \binom{n+3}{3}$

   2. Plug in the given indices

   $\alpha_{12} = \binom{12+4}{4} = \binom{16}{4}$

   $\beta_{13} = \binom{13+3}{3} = \binom{16}{3}$

   3. Apply Pascal’s identity Pascal’s rule: $\binom{m}{r} + \binom{m}{r-1} = \binom{m+1}{r}$. Here $m=16$, $r=4$: $$$$

   🔒 Start your 14-day free trial to see the full solution →
2. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If $x = 2 + \dfrac{7}{8} + \dfrac{7 \cdot 10}{8 \cdot 12} + \dfrac{7 \cdot 10 \cdot 13}{8 \cdot 12 \cdot 16} + \ldots \infty$, then $x^3 =$ (A) $81$ (B) $625$ (C) $256$ (D) $216$

   ›Reveal solutionSolution

   The series is a binomial-type expansion whose sum is $x = 2^{8/3}$, hence $x^3 = 2^{8} = 256$ — option (C).

   Write the series with its "binomial" part visible:

   $x = 2 + \frac{7}{8} + \frac{7\cdot 10}{8\cdot 12} + \frac{7\cdot 10\cdot 13}{8\cdot 12\cdot 16} + \cdots$

   Ratio of consecutive terms. For the $k$-th term past the leading $2$, the multiplying factor is $\dfrac{3k+4}{4(k+1)}$, since the numerators step by $3$ from $7$ and the denominators step by $4$ from $8$.

   Closed form. Let $S = 1 + \dfrac{7}{8} + \dfrac{7\cdot 10}{8\cdot 12}+\cdots$ (the same series but with leading term $1$). Its general term is

   $c_k = \frac{\left(\tfrac{7}{3}\right)_k}{(k+1)!}\left(\tfrac{3}{4}\right)^k,\qquad\text{so}\qquad S=\frac{1}{z}\int_0^{z}(1-t)^{-7/3}\,dt,\quad z=\tfrac34.$

   Evaluate the integral.

   $\int_0^{z}(1-t)^{-7/3}\,dt=\frac{3}{4}\Big[(1-z)^{-4/3}-1\Big].$ …

   🔒 Start your 14-day free trial to see the full solution →
3. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If $(\sin\theta + \cos\theta)^4 + (\sin\theta - \cos\theta)^4 = p - q(\sin^4\theta + \cos^4\theta)$, then $p + q =$ (A) $6$ (B) $4$ (C) $10$ (D) $8$

   ›Reveal solutionSolution

   The key is to expand both fourth-power binomials, simplify using $\sin^2\theta + \cos^2\theta = 1$, and match coefficients to find $p$ and $q$, yielding $p+q = 8$.

   We start with the given expression:

   $(\sin\theta + \cos\theta)^4 + (\sin\theta - \cos\theta)^4 = p - q(\sin^4\theta + \cos^4\theta)$

   Our goal is to find $p+q$.

   ---

   Concept & Intuition

   When you see symmetric binomials like $(a+b)^4 + (a-b)^4$, the cross terms cancel in a neat way: the odd powers of $b$ vanish, leaving only even powers. This reduces the algebra dramatically. Then, using the Pythagorean identity, we can express everything in terms of $\sin^4\theta + \cos^4\theta$, allowing us to read off $p$ and $q$ directly.

   ---

   Step-by-step solution

   1. Expand each binomial Recall $(x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$. Let $a = \sin\theta$, $b = \cos\theta$. Then:

   $(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$

   $(a-b)^4 = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4$

   2. Add them The terms with odd powers of $b$ ($4a^3b$ and $4ab^3$) cancel:

   $(a+b)^4 + (a-b)^4 = 2a^4 + 12a^2b^2 + 2b^4$

   3. Factor and rewrite Factor 2:

   $= 2(a^4 + b^4 + 6a^2b^2)$

   But $a^2b^2 = \sin^2\theta \cos^2\theta$. We can relate this to $\sin^4\theta + \cos^4\theta$ using:

   $(\sin^2\theta + \cos^2\theta)^2 = 1 = \sin^4\theta + \cos^4\theta + 2\sin^2\theta\cos^2\theta$

   So:

   $\sin^2\theta\cos^2\theta = \frac{1 - (\sin^4\theta + \cos^4\theta)}{2}$

   4. Substitute back

   $\text{LHS} = 2\left[ \sin^4\theta + \cos^4\theta + 6\cdot\frac{1 - (\sin^4\theta + \cos^4\theta)}{2} \right]$

   Simplify inside:

   $= 2\left[ \sin^4\theta + \cos^4\theta + 3 - 3(\sin^4\theta + \cos^4\theta) \right]$

   $= 2\left[ 3 - 2(\sin^4\theta + \cos^4\theta) \right]$

   $$$$

   🔒 Start your 14-day free trial to see the full solution →
4. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If $|x|$ is so small that $x^2$ and higher powers of $x$ may be neglected and hence

   [!FORMULA] $\frac{(8-3x)^{1/3}}{(2+3x)^{1/3}} \approx \frac{1}{a}(1+bx)$

   then $2ab=$ (A) $-\frac{37}{2}$ (B) $-\frac{35}{2}$ (C) $-37$ (D) $-35$

   ›Reveal solutionSolution

   Binomial expansion to first order gives $\frac14\left(1-\frac{37x}{8}\right)$, so $a=4$ and $b=-\frac{37}{8}$, whence $2ab=\mathbf{-37}$ — option (C).

   The concept first

   The binomial theorem for any index $n$ (not just positive integers) says that for $|u|<1$,

   $(1+u)^{n}=1+nu+\frac{n(n-1)}{2!}u^{2}+\cdots$

   When we are told that $x^2$ and higher powers may be neglected, this collapses to the workhorse approximation

   $\boxed{(1+u)^{n}\approx 1+nu}$

   Two practical rules make these questions painless:

   1. Always factor the constant out first so the bracket looks like $(1+\text{small})^n$ — you cannot expand $(8-3x)^{1/3}$ directly, but you can expand $\left(1-\frac{3x}{8}\right)^{1/3}$.
   2. Turn division into a negative index: $\dfrac{1}{(1+u)^{m}}=(1+u)^{-m}\approx 1-mu$.

   Step-by-step

   Step 1 — factor the constants out.

   $(8-3x)^{1/3}=\left[8\left(1-\frac{3x}{8}\right)\right]^{1/3}=8^{1/3}\left(1-\frac{3x}{8}\right)^{1/3}=2\left(1-\frac{3x}{8}\right)^{1/3}$

   $(2+3x)^{3}=\left[2\left(1+\frac{3x}{2}\right)\right]^{3}=8\left(1+\frac{3x}{2}\right)^{3}$

   Step 2 — expand each bracket to first order.

   $\left(1-\frac{3x}{8}\right)^{1/3}\approx 1+\frac13\left(-\frac{3x}{8}\right)=1-\frac{x}{8}$

   $\left(1+\frac{3x}{2}\right)^{-3}\approx 1+(-3)\left(\frac{3x}{2}\right)=1-\frac{9x}{2}$

   Step 3 — combine.

   $\frac{(8-3x)^{1/3}}{(2+3x)^{3}}\approx\frac{2}{8}\left(1-\frac{x}{8}\right)\left(1-\frac{9x}{2}\right)=\frac14\left(1-\frac{x}{8}-\frac{9x}{2}+\underbrace{\frac{9x^{2}}{16}}_{\text{neglect}}\right)$

   Step 4 — collect the $x$ terms.

   $-\frac{x}{8}-\frac{9x}{2}=-\frac{x}{8}-\frac{36x}{8}=-\frac{37x}{8}$

   so …

   🔒 Start your 14-day free trial to see the full solution →
5. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.The coefficient of $x^2$ in the power series expansion of $\dfrac{2x}{(x-1)(x+2)^2}$ when $|x|<1$ is (A) $0$ (B) $-\dfrac{2}{3}$ (C) $\dfrac{4}{9}$ (D) $\dfrac{1}{3}$

   ›Reveal solutionSolution

   The key is to decompose the rational function into partial fractions, then expand each term as a power series using the binomial theorem for negative exponents. The coefficient of $x^2$ turns out to be $-\frac{2}{3}$.

   We are asked for the coefficient of $x^2$ in the power series expansion of $\dfrac{2x}{(x-1)(x+2)^2}$ when $|x|<1$. The condition $|x|<1$ tells us we are expanding about $x=0$ in a convergent series, and it also tells us which form of expansion to use for each factor.

   The direct approach — trying to expand the whole fraction as a single series — is messy. Instead, we break the rational function into simpler pieces whose series expansions we know. This is the method of partial fractions, and it is the natural tool here because the denominator factors nicely.

   ---

   1. Set up the partial fraction decomposition

   We have

   $\frac{2x}{(x-1)(x+2)^2}.$

   Since the denominator has a linear factor $(x-1)$ and a repeated linear factor $(x+2)^2$, the partial fraction form is

   $\frac{2x}{(x-1)(x+2)^2} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{(x+2)^2}.$

   Multiply through by $(x-1)(x+2)^2$:

   $2x = A(x+2)^2 + B(x-1)(x+2) + C(x-1).$

   ---

   2. Solve for the constants

   We can find $A$ and $C$ quickly by substituting convenient values of $x$.

   For $A$: Set $x = 1$. Then $(x-1)=0$, so the $B$ and $C$ terms vanish:

   $2(1) = A(1+2)^2 \quad\Rightarrow\quad 2 = 9A \quad\Rightarrow\quad A = \frac{2}{9}.$

   For $C$: Set $x = -2$. Then $(x+2)=0$, so the $A$ and $B$ terms vanish:

   $2(-2) = C(-2-1) \quad\Rightarrow\quad -4 = -3C \quad\Rightarrow\quad C = \frac{4}{3}.$

   Now find $B$ by substituting any other value, say $x = 0$:

   $0 = A(2)^2 + B(-1)(2) + C(-1) = 4A - 2B - C.$

   Plug $A = \frac{2}{9}$ and $C = \frac{4}{3}$:

   $0 = 4\left(\frac{2}{9}\right) - 2B - \frac{4}{3} = \frac{8}{9} - 2B - \frac{12}{9} = -\frac{4}{9} - 2B.$

   Thus $2B = -\frac{4}{9}$, so $B = -\frac{2}{9}$.

   So we have

   $\frac{2x}{(x-1)(x+2)^2} = \frac{2/9}{x-1} - \frac{2/9}{x+2} + \frac{4/3}{(x+2)^2}.$

   ---

   3. Rewrite each term for expansion about $x=0$

   We want series in powers of $x$, so we write each denominator in the form $(\text{constant})(1 \pm \text{something in }x)$.

   For $\frac{2/9}{x-1}$:

   $\frac{2/9}{x-1} = -\frac{2/9}{1-x} = -\frac{2}{9} \cdot \frac{1}{1-x}.$

   For $-\frac{2/9}{x+2}$:

   $-\frac{2/9}{x+2} = -\frac{2}{9} \cdot \frac{1}{2+x} = -\frac{2}{9} \cdot \frac{1}{2(1 + x/2)} = -\frac{1}{9} \cdot \frac{1}{1 + x/2}.$

   For $\frac{4/3}{(x+2)^2}$:

   $\frac{4/3}{(x+2)^2} = \frac{4}{3} \cdot \frac{1}{(2+x)^2} = \frac{4}{3} \cdot \frac{1}{4(1 + x/2)^2} = \frac{1}{3} \cdot \frac{1}{(1 + x/2)^2}.$

   Thus the expression becomes

   $-\frac{2}{9}\cdot\frac{1}{1-x} \;-\; \frac{1}{9}\cdot\frac{1}{1 + x/2} \;+\; \frac{1}{3}\cdot\frac{1}{(1 + x/2)^2}.$

   ---

   4. Expand each using known series

   For $|x|<1$, we have the geometric series:

   $\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots.$ …

   🔒 Start your 14-day free trial to see the full solution →
6. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.The approximate value of $(0.98)^{0.2}$, rounded to 4 decimal places, found by using binomial expansion is (A) 0.9860 (B) 0.9950 (C) 0.9960 (D) 1.0060

   ›Reveal solutionSolution

   To approximate $(0.98)^{0.2}$, we rewrite it as $(1 - 0.02)^{0.2}$ and use the binomial expansion $(1+x)^n \approx 1 + nx$ for small $x$. This yields an approximate value of $\boxed{0.9960}$.

   When we need to find the approximate value of an expression like $(0.98)^{0.2}$, especially when the base is close to 1, the binomial expansion is a powerful tool. The core idea is to transform the expression into the form $(1+x)^n$, where $x$ is a small number.

   The binomial expansion for any real number $n$ and for $|x| < 1$ is given by:

   $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots$

   The utility of this expansion for approximation comes from the fact that if $x$ is a small number (i.e., close to zero), then $x^2$ will be much smaller than $x$, $x^3$ will be much smaller than $x^2$, and so on. This means that the terms in the series decrease rapidly in magnitude. For a good approximation, we often only need to consider the first few terms, typically $1 + nx$, or sometimes $1 + nx + \frac{n(n-1)}{2!}x^2$, depending on the required precision.

   Let's apply this to the given problem.

   1. Rewrite the expression in the form $(1+x)^n$.

      The given expression is $(0.98)^{0.2}$. We can write $0.98$ as $1 - 0.02$.

      So, $(0.98)^{0.2} = (1 - 0.02)^{0.2}$.

      Comparing this with $(1+x)^n$, we identify $x = -0.02$ and $n = 0.2$.

      Notice that $x = -0.02$ is indeed a small number, which makes the binomial expansion suitable for approximation.

   2. Apply the binomial expansion formula.

      Using the formula $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots$, we substitute $x = -0.02$ and $n = 0.2$:

   $(1 - 0.02)^{0.2} = 1 + (0.2)(-0.02) + \frac{(0.2)(0.2-1)}{2!}(-0.02)^2 + \dots$

   3. Calculate the first few terms.

      • The first term is $1$.
      • The second term is $nx = (0.2)(-0.02) = -0.004$.
      • The third term is $\frac{n(n-1)}{2!}x^2 = \frac{(0.2)(-0.8)}{2}(0.0004) = \frac{-0.16}{2}(0.0004) = -0.08(0.0004) = -0.000032$.

   4. Sum the terms and round to the required precision.

      Now, we sum these terms to get the approximate value: …

   🔒 Start your 14-day free trial to see the full solution →
7. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If $\dfrac{2x^6 + 3x^4 + 1}{(x^2 + 2)^4} = \dfrac{Ax + P}{x^2 + 2} + \dfrac{Bx + Q}{(x^2 + 2)^2} + \dfrac{Cx + R}{(x^2 + 2)^3} + \dfrac{Dx + T}{(x^2 + 2)^4}$, then $3P + 2Q + R + 4T =$ (A) $-12$ (B) $30$ (C) $-3$ (D) $24$

   ›Reveal solutionSolution

   By substituting $y = x^2 + 2$, the given rational function simplifies to a direct sum of terms, allowing us to easily identify the coefficients $P, Q, R, T$. The final value is $\boxed{-12}$.

   The problem asks us to find the value of an expression involving coefficients $P, Q, R, T$ from a partial fraction decomposition. The given rational function has a denominator that is a power of an irreducible quadratic factor, $(x^2 + 2)^4$. Notice that the numerator, $2x^6 + 3x^4 + 1$, is a polynomial where all powers of $x$ are even. This is a crucial observation.

   This structure suggests a powerful simplification: we can treat $x^2$ as a variable. If we let $y = x^2 + 2$, then $x^2 = y - 2$. Substituting this into the numerator will transform the entire expression into a rational function of $y$, which will be much easier to decompose. This method works because the numerator is purely a function of $x^2$, meaning it will not introduce any odd powers of $x$ that would lead to non-zero $A, B, C, D$ coefficients in the partial fraction form.

   1. Substitute $y = x^2 + 2$:

      Let $y = x^2 + 2$. This implies $x^2 = y - 2$. We will rewrite the numerator in terms of $y$.

   2. Rewrite the numerator in terms of $y$:

      The numerator is $2x^6 + 3x^4 + 1$.

      Substitute $x^2 = y - 2$:

      $2(x^2)^3 + 3(x^2)^2 + 1 = 2(y - 2)^3 + 3(y - 2)^2 + 1$.

      Now, expand the terms:

      $2(y^3 - 3y^2(2) + 3y(2^2) - 2^3) + 3(y^2 - 2y(2) + 2^2) + 1$

      $= 2(y^3 - 6y^2 + 12y - 8) + 3(y^2 - 4y + 4) + 1$

      $= 2y^3 - 12y^2 + 24y - 16 + 3y^2 - 12y + 12 + 1$

      Combine like terms:

      $= 2y^3 + (-12 + 3)y^2 + (24 - 12)y + (-16 + 12 + 1)$

      $= 2y^3 - 9y^2 + 12y - 3$.

   3. Rewrite the entire fraction in terms of $y$:

      The original expression is $\dfrac{2x^6 + 3x^4 + 1}{(x^2 + 2)^4}$.

      Using our substitution, this becomes:

      $\dfrac{2y^3 - 9y^2 + 12y - 3}{y^4}$.

   4. Simplify the fraction:

      We can split this into individual terms by dividing each term in the numerator by $y^4$:

      $\dfrac{2y^3}{y^4} - \dfrac{9y^2}{y^4} + \dfrac{12y}{y^4} - \dfrac{3}{y^4}$

      $= \dfrac{2}{y} - \dfrac{9}{y^2} + \dfrac{12}{y^3} - \dfrac{3}{y^4}$.

   5. Substitute back $x^2 + 2$ for $y$:

      Now, replace $y$ with $x^2 + 2$:

      $\dfrac{2}{x^2 + 2} - \dfrac{9}{(x^2 + 2)^2} + \dfrac{12}{(x^2 + 2)^3} - \dfrac{3}{(x^2 + 2)^4}$.

   6. Compare with the given partial fraction form:

      The given form is $\dfrac{Ax + P}{x^2 + 2} + \dfrac{Bx + Q}{(x^2 + 2)^2} + \dfrac{Cx + R}{(x^2 + 2)^3} + \dfrac{Dx + T}{(x^2 + 2)^4}$.

      By comparing our simplified expression with this form, we can identify the coefficients:

      • For the term with denominator $(x^2 + 2)$: $\dfrac{Ax + P}{x^2 + 2} = \dfrac{2}{x^2 + 2} \implies Ax + P = 2$. This means $A = 0$ and $P = 2$. …

   🔒 Start your 14-day free trial to see the full solution →
8. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.In the binomial expansion of $(1+x)^n$, the coefficients of $x^{k-1}, x^k, x^{k+1}$ and also the coefficients of $x^{l-1}, x^l, x^{l+1}$ are in A.P. Then numerically greatest term in the expansion of $(1+x)^{l+k}$ when $x = \dfrac{2}{3}$ and $n = 14$ is (A) ${}^{14}C_5 \left(\dfrac{2}{3}\right)^5$ (B) ${}^{14}C_4 \left(\dfrac{2}{3}\right)^4$ (C) ${}^{15}C_7 \left(\dfrac{2}{3}\right)^8$ (D) ${}^{15}C_6 \left(\dfrac{2}{3}\right)^9$

   ›Reveal solutionSolution

   The problem uses the condition that three consecutive binomial coefficients are in arithmetic progression to find integer parameters $k$ and $l$, then identifies the numerically greatest term in $(1+x)^{l+k}$ for $x=2/3$ and $n=14$, leading to option (C).

   We start with the binomial expansion of $(1+x)^n$:

   $(1+x)^n = \sum_{r=0}^n \binom{n}{r} x^r.$

   The coefficients of $x^{k-1}, x^k, x^{k+1}$ are $\binom{n}{k-1}, \binom{n}{k}, \binom{n}{k+1}$. The problem says these three are in arithmetic progression (AP). The same holds for another triple with index $l$. This gives us equations to solve for $k$ and $l$ in terms of $n$. Then we use $n=14$ and $x=2/3$ to find the numerically greatest term in $(1+x)^{l+k}$.

   ---

   1. Condition for three consecutive binomial coefficients to be in AP

   For three numbers $a, b, c$ to be in AP, we have $2b = a + c$. So:

   $2\binom{n}{k} = \binom{n}{k-1} + \binom{n}{k+1}.$

   Recall the factorial form:

   $\binom{n}{k} = \frac{n!}{k!(n-k)!}, \quad \binom{n}{k-1} = \frac{n!}{(k-1)!(n-k+1)!}, \quad \binom{n}{k+1} = \frac{n!}{(k+1)!(n-k-1)!}.$

   Divide the whole equation by $\frac{n!}{(k-1)!(n-k-1)!}$ to simplify. Let’s do it step by step.

   First, write:

   $2 \cdot \frac{n!}{k!(n-k)!} = \frac{n!}{(k-1)!(n-k+1)!} + \frac{n!}{(k+1)!(n-k-1)!}.$

   Cancel $n!$:

   $\frac{2}{k!(n-k)!} = \frac{1}{(k-1)!(n-k+1)!} + \frac{1}{(k+1)!(n-k-1)!}.$

   Multiply through by $k!(n-k)!$:

   $2 = \frac{k!(n-k)!}{(k-1)!(n-k+1)!} + \frac{k!(n-k)!}{(k+1)!(n-k-1)!}.$

   Simplify each fraction:

   • First term: $\frac{k!}{(k-1)!} = k$, $\frac{(n-k)!}{(n-k+1)!} = \frac{1}{n-k+1}$, so first term = $\frac{k}{n-k+1}$.
   • Second term: $\frac{k!}{(k+1)!} = \frac{1}{k+1}$, $\frac{(n-k)!}{(n-k-1)!} = n-k$, so second term = $\frac{n-k}{k+1}$.

   Thus:

   $\frac{k}{n-k+1} + \frac{n-k}{k+1} = 2.$

   2. Solve the equation

   Multiply both sides by $(n-k+1)(k+1)$:

   $k(k+1) + (n-k)(n-k+1) = 2(n-k+1)(k+1).$

   Expand:

   • Left: $k^2 + k + (n-k)(n-k+1) = k^2 + k + (n^2 - nk + n - nk + k^2 - k) = k^2 + k + n^2 - 2nk + n + k^2 - k$. Simplify: $2k^2 + n^2 - 2nk + n$.
   • Right: $2[(n-k+1)(k+1)] = 2[(n-k+1)k + (n-k+1)] = 2[nk - k^2 + k + n - k + 1] = 2[nk - k^2 + n + 1]$.

   So equation:

   $2k^2 + n^2 - 2nk + n = 2nk - 2k^2 + 2n + 2.$

   Bring all terms to one side:

   $2k^2 + n^2 - 2nk + n - 2nk + 2k^2 - 2n - 2 = 0,$

   $4k^2 - 4nk + n^2 - n - 2 = 0.$

   Divide by 1:

   $4k^2 - 4nk + (n^2 - n - 2) = 0.$

   This is quadratic in $k$:

   $k = \frac{4n \pm \sqrt{16n^2 - 16(n^2 - n - 2)}}{8} = \frac{4n \pm \sqrt{16n + 32}}{8} = \frac{4n \pm 4\sqrt{n+2}}{8} = \frac{n \pm \sqrt{n+2}}{2}.$

   So the two possible integer values for $k$ (and similarly for $l$) are:

   $k = \frac{n - \sqrt{n+2}}{2}, \quad \text{or} \quad k = \frac{n + \sqrt{n+2}}{2}.$

   For these to be integers, $n+2$ must be a perfect square. With $n=14$, $n+2=16$, so $\sqrt{16}=4$. Then:

   $k = \frac{14 - 4}{2} = 5, \quad \text{or} \quad k = \frac{14 + 4}{2} = 9.$

   Thus the two indices are $k=5$ and $l=9$ (or vice versa). So $l+k = 14$.

   Tip

   The two solutions are symmetric about $n/2$; they are the two positions where three consecutive binomial coefficients are in AP. For $n=14$, these are $r=5$ and $r=9$.

   3. The expansion now is $(1+x)^{14}$ with $x = 2/3$

   We need the numerically greatest term in $(1 + \frac{2}{3})^{14}$. For a binomial $(1+ax)^n$ with $a>0$, the numerically greatest term occurs at the integer $r$ satisfying: …

   🔒 Start your 14-day free trial to see the full solution →
9. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If $\dfrac{x^3+3}{(x-3)^3}=a+\dfrac{b}{x-3}+\dfrac{c}{(x-3)^2}+\dfrac{d}{(x-3)^3}$ then $(a+d)-(b+c)=$ (A) $49$ (B) $15$ (C) $-30$ (D) $-5$

   ›Reveal solutionSolution

   We rewrite the rational expression by performing a substitution and polynomial expansion, then match coefficients to find $a,b,c,d$ and compute $(a+d)-(b+c) = 49$. The correct option is (A).

   ---

   We are given

   $\frac{x^3+3}{(x-3)^3}=a+\frac{b}{x-3}+\frac{c}{(x-3)^2}+\frac{d}{(x-3)^3}$

   and need $(a+d)-(b+c)$.

   Concept & Intuition

   The right-hand side is the partial fraction decomposition of the left-hand side, but with a twist: the numerator degree (3) equals the denominator degree (3), so there is a constant term $a$ (the “polynomial part”). The usual approach is to clear denominators and equate coefficients, but that can be messy. A cleaner method: let $t = x-3$, so $x = t+3$. Then the left side becomes a rational function in $t$, which we expand as a sum of powers of $t$. Matching terms to the right side (now in $t$) gives the constants directly.

   ---

   Step-by-step solution

   1. Substitute $t = x-3$ Then $x = t+3$. The left side becomes

   $\frac{(t+3)^3 + 3}{t^3}.$

   2. Expand the numerator

   $(t+3)^3 = t^3 + 9t^2 + 27t + 27.$

   Adding 3:

   $t^3 + 9t^2 + 27t + 30.$

   So

   $\frac{x^3+3}{(x-3)^3} = \frac{t^3 + 9t^2 + 27t + 30}{t^3}.$

   3. Separate into partial fractions in $t$

   $\frac{t^3 + 9t^2 + 27t + 30}{t^3} = 1 + \frac{9}{t} + \frac{27}{t^2} + \frac{30}{t^3}.$

   4. Translate back to $x$ Since $t = x-3$,

   $\frac{x^3+3}{(x-3)^3} = 1 + \frac{9}{x-3} + \frac{27}{(x-3)^2} + \frac{30}{(x-3)^3}.$

   5. Match coefficients Comparing with $$$$

   🔒 Start your 14-day free trial to see the full solution →
10. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.If $C_0, C_1, C_2, ..., C_n$ are the binomial coefficients in the expansion of $(1+x)^n$ then the value of $\sum r^3 \cdot C_r$ when $n=5$ is (A) 320 (B) 560 (C) 720 (D) 800

    ›Reveal solutionSolution

    $\sum_{r=0}^{5} r^3\binom{5}{r}=800$, so the answer is (D).

    Evaluate directly for $n=5$ using $\binom{5}{r}=1,5,10,10,5,1$:

    $\sum_{r=0}^{5} r^{3}\binom{5}{r}=0+1^3(5)+2^3(10)+3^3(10)+4^3(5)+5^3(1).$

    $=0+5+80+270+320+125=800.$ …

    🔒 Start your 14-day free trial to see the full solution →
11. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
    Q.If $C_0, C_1, C_2, ..., C_n$ are the binomial coefficients in the expansion of $(1+x)^n$ then the value of $\sum r^3 \cdot C_r$ when $n=5$ is (A) 800 (B) 320 (C) 560 (D) 720

    ›Reveal solutionSolution

    The sum $\sum_{r=0}^n r^3 \binom{n}{r}$ can be evaluated using combinatorial identities or by differentiating the binomial expansion. For $n=5$, the result is $720$, which corresponds to option (D).

    We want $\sum_{r=0}^5 r^3 \binom{5}{r}$. The key idea is that sums of the form $\sum r^k \binom{n}{r}$ are related to moments of the binomial distribution, and can be computed elegantly using the binomial theorem and its derivatives, or by using known combinatorial identities like $r\binom{n}{r} = n\binom{n-1}{r-1}$. This avoids brute-force summing all terms.

    ---

    Step-by-step solution:

    1. Recall the identity for $r \binom{n}{r}$:

    $r \binom{n}{r} = n \binom{n-1}{r-1}$

    This holds for $r \ge 1$. It lets us reduce the power of $r$ by one, at the cost of lowering $n$.

    2. Write $r^3 = r \cdot r^2$ and apply the identity twice: First,

    $r^3 \binom{n}{r} = r \cdot r^2 \binom{n}{r} = r \cdot \left[ n \binom{n-1}{r-1} \cdot r \right]?$

    Careful: we need to handle $r^2$ inside. Better: use

    $r^3 \binom{n}{r} = r \cdot \left( r^2 \binom{n}{r} \right)$

    But a more systematic method is to express $r^3$ as a linear combination of falling factorials:

    $r^3 = r(r-1)(r-2) + 3r(r-1) + r$

    because falling factorials match nicely with binomial coefficients.

    3. Verify the decomposition:

    $r(r-1)(r-2) = r^3 - 3r^2 + 2r$

    $3r(r-1) = 3r^2 - 3r$

    Adding: $r^3 - 3r^2 + 2r + 3r^2 - 3r + r = r^3$. Yes.

    4. Now sum term by term:

    $\sum_{r=0}^n r^3 \binom{n}{r} = \sum_{r=0}^n r(r-1)(r-2) \binom{n}{r} + 3\sum_{r=0}^n r(r-1) \binom{n}{r} + \sum_{r=0}^n r \binom{n}{r}$

    5. Use the falling factorial identity: For $k \le n$,

    $r(r-1)\cdots(r-k+1) \binom{n}{r} = n(n-1)\cdots(n-k+1) \binom{n-k}{r-k}$

    This holds for $r \ge k$ and is zero otherwise. So:

    • For $k=3$:

    $\sum_{r=0}^n r(r-1)(r-2) \binom{n}{r} = n(n-1)(n-2) \sum_{r=3}^n \binom{n-3}{r-3} = n(n-1)(n-2) \cdot 2^{n-3}$

    • For $k=2$:

    $\sum_{r=0}^n r(r-1) \binom{n}{r} = n(n-1) \cdot 2^{n-2}$

    • For $k=1$:

    $\sum_{r=0}^n r \binom{n}{r} = n \cdot 2^{n-1}$

    6. Plug in $n=5$:

    $\sum r^3 \binom{5}{r} = 5\cdot4\cdot3 \cdot 2^{2} + 3 \cdot (5\cdot4 \cdot 2^{3}) + (5 \cdot 2^{4})$

    Compute each:

    • $5\cdot4\cdot3 = 60$, $2^{2}=4$ → $60 \times 4 = 240$
    • $5\cdot4 = 20$, $2^{3}=8$ → $20 \times 8 = 160$, times 3 gives $480$
    • $5 \cdot 2^{4} = 5 \times 16 = 80$

    Sum: $240 + 480 + 80 = 800$.

    7. Wait — that gives 800, but let’s double-check: …

    🔒 Start your 14-day free trial to see the full solution →
12. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.The number of ways in which 4 different things can be distributed to 6 persons so that no person gets all the things is (A) 1292 (B) 1296 (C) 1290 (D) 4090

    ›Reveal solutionSolution

    We count all distributions of 4 distinct items to 6 distinct people, then subtract the forbidden cases where someone gets all 4 items. The result is $6^4 - 6 = 1296 - 6 = 1290$, so the answer is (C).

    Concept & Intuition

    When distributing different things to different people, each item has an independent choice of recipient. That gives a total of (\text{#people})^{\text{#items}} ways. The only restriction here is that no single person receives all the items. That’s a classic “total minus forbidden” situation: count everything, then subtract the few cases that violate the rule. The forbidden cases are easy: pick which person gets everything (6 choices), and then there’s exactly 1 way to give them all 4 items. No other restrictions apply, so the subtraction is clean.

    Step-by-step

    1. Total distributions without restriction Each of the 4 different things can go to any of the 6 persons independently.

    $\text{Total} = 6^4 = 1296.$

    2. Identify the forbidden distributions The only forbidden outcome is when one person receives all 4 things.
       • Choose the person who gets everything: $6$ ways.
       • Give them all 4 items: only $1$ way (since the items are distinct, but they all go to the same person). So the number of forbidden distributions is $6 \times 1 = 6.$ …

    🔒 Start your 14-day free trial to see the full solution →