Q.Compute $(98)^5$.

Showing the 12 most recent of 19 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.If the coefficients of $x^2$ and $x^3$ in the expansion of $(3+k x)^9$ are equal, then the value of ' $\boldsymbol{k}$ ' is (A) $\frac{7}{3}$ (B) $\frac{7}{9}$ (C) $\frac{9}{7}$ (D) $\frac{3}{7}$

   ›Reveal solutionSolution

   The problem uses the binomial theorem to equate the coefficients of $x^2$ and $x^3$ in $(3+kx)^9$, leading to a simple equation that solves to $k = \frac{9}{7}$, which corresponds to option (C).

   We are expanding $(3 + kx)^9$. The binomial theorem tells us that the general term is $\binom{9}{r} 3^{9-r} (kx)^r$. The coefficient of $x^r$ is $\binom{9}{r} 3^{9-r} k^r$.

   The key idea: set the coefficient of $x^2$ equal to the coefficient of $x^3$, then solve for $k$. This works because the problem states they are equal — no extra conditions.

   1. Write the coefficient of $x^2$

      For $r=2$: $\binom{9}{2} 3^{9-2} k^2 = \binom{9}{2} 3^{7} k^2$.

   2. Write the coefficient of $x^3$

      For $r=3$: $\binom{9}{3} 3^{9-3} k^3 = \binom{9}{3} 3^{6} k^3$.

   3. Set them equal

   $\binom{9}{2} 3^{7} k^2 = \binom{9}{3} 3^{6} k^3$

   4. Simplify the binomial coefficients $\binom{9}{2} = 36$ and $\binom{9}{3} = 84$. So the equation becomes:

   $36 \cdot 3^{7} k^2 = 84 \cdot 3^{6} k^3$

   5. Cancel common factors Divide both sides by $3^{6} k^2$ (assuming $k \neq 0$, which is fine since $k=0$ would make both coefficients zero but then they'd be equal trivially — but the options suggest a nonzero answer):

   $36 \cdot 3 = 84 \cdot k$

   That is:

   $108 = 84k$

   6. Solve for $k$

   $k = \frac{108}{84} = \frac{9}{7}$

   Tip

   Notice we didn't need to compute $3^7$ or $3^6$ fully — just the ratio $3^7 / 3^6 = 3$ simplifies nicely. Always look for such cancellations.

   Watch out

   A common mistake is forgetting that the coefficient includes the power of 3 as well as the binomial coefficient. If you only equate $\binom{9}{2} k^2 = \binom{9}{3} k^3$, you'd get $k = \frac{2}{3}$, which is not among the options — a sign you missed the 3's.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

2. COMEDK 2026Set 2026-M1 markMCQ
   Q.\text { The remainder when } \mathbf{7}^{\mathbf{1 0 3}} \text { is divided by } \mathbf{2 5} \text { is } (A) 7 (B) 18 (C) 1 (D) -1

   ›Reveal solutionSolution

   We use Euler’s theorem to reduce the exponent modulo φ(25)=20, then compute the remainder of 7^103 mod 25. The remainder is 18, so the correct option is (B).

   We want $7^{103} \bmod 25$. Direct computation is impossible, so we need a clever reduction. The key idea: since 7 and 25 are coprime, Euler’s theorem tells us that $7^{\phi(25)} \equiv 1 \pmod{25}$. This lets us reduce the huge exponent 103 to a much smaller one.

   1. Compute Euler’s totient for 25

      $25 = 5^2$, so $\phi(25) = 25 \cdot (1 - \frac{1}{5}) = 25 \cdot \frac{4}{5} = 20$.

      Thus $7^{20} \equiv 1 \pmod{25}$.

   2. Reduce the exponent modulo 20

      We write $103 = 20 \cdot 5 + 3$.

      Then $7^{103} = 7^{20 \cdot 5 + 3} = (7^{20})^5 \cdot 7^3 \equiv 1^5 \cdot 7^3 \pmod{25}$.

      So the problem reduces to finding $7^3 \bmod 25$.

   3. Compute $7^3$ and its remainder

      $7^3 = 343$. Now divide 343 by 25:

      $25 \times 13 = 325$, remainder $343 - 325 = 18$.

      So $7^3 \equiv 18 \pmod{25}$.

   4. Conclusion

      Therefore $7^{103} \equiv 18 \pmod{25}$. The remainder is 18.

   Watch out

   A common mistake is to forget that Euler’s theorem requires the base and modulus to be coprime. Here 7 and 25 are coprime, so it’s safe. If they weren’t, we’d need a different method (like the Chinese remainder theorem).

   Tip

   Notice that 18 is also $-7$ modulo 25, which might appear in some answer choices as “-1” if miscomputed. Always reduce to a positive remainder between 0 and 24.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

3. COMEDK 2025Set 2025-A1 markMCQ
   Q.Evaluate the value of $(1.02)^8$ using binomial theorem up to two decimal places. (A) 1.17 (B) 1.18 (C) 1.81 (D) 1.71

   ›Reveal solutionSolution

   Using the binomial expansion $(1+x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2$ with $x=0.02$ and $n=8$, we get $1.17$ correct to two decimal places, matching option (A).

   We want $(1.02)^8$ without a calculator, accurate to two decimal places. The binomial theorem lets us expand $(1+x)^n$ as a sum of terms, and for small $x$, the first few terms give an excellent approximation. Here $x = 0.02$ is small, so we can truncate after the $x^2$ term — the error will be tiny.

   1. Write the expression in binomial form $(1.02)^8 = (1 + 0.02)^8$. The binomial theorem says:

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

   2. Decide how many terms we need

      For two decimal place accuracy, we need the result correct to the nearest hundredth.

      • $x = 0.02$, so $x^2 = 0.0004$, $x^3 = 0.000008$, etc.
      • The $x^2$ term will contribute about $\frac{8 \cdot 7}{2} \cdot 0.0004 = 28 \cdot 0.0004 = 0.0112$, which affects the second decimal place.
      • The $x^3$ term is $\frac{8 \cdot 7 \cdot 6}{6} \cdot 0.000008 = 56 \cdot 0.000008 = 0.000448$, which only changes the third decimal place. So we can safely stop after the $x^2$ term.

   3. Compute the first three terms

      • First term: $1$
      • Second term: $nx = 8 \times 0.02 = 0.16$
      • Third term: $\frac{n(n-1)}{2}x^2 = \frac{8 \times 7}{2} \times (0.02)^2 = 28 \times 0.0004 = 0.0112$

   4. Add them up

   $1 + 0.16 + 0.0112 = 1.1712$

   5. Round to two decimal places $1.1712$ rounds to $1.17$ (since the third decimal is $1$, less than $5$).

   Watch out

   A common mistake is to forget the $x^2$ term and just compute $1 + 0.16 = 1.16$, which would be off by $0.01$ — enough to pick the wrong option. Always check whether the quadratic term affects the required precision.

   Tip

   For $(1+x)^n$ with small $x$, the approximation $1 + nx$ is linear, but adding the quadratic term $\frac{n(n-1)}{2}x^2$ often gives the exact two-decimal-place answer. Here it lifts $1.16$ to $1.17$.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

4. COMEDK 2025Set 2025-M1 markMCQ
   Q.If the third and fourth terms in the expansion $\left(2 x+\frac{1}{8}\right)^{10}$ are equal, then the value of $x$ is __________ (A) $\frac{1}{8}$ (B) $\frac{2}{7}$ (C) $\frac{1}{6}$ (D) $\frac{8}{3}$

   ›Reveal solutionSolution

   The key idea is to equate the third and fourth terms of the binomial expansion using the general term formula, then solve for $x$. The value is $x = \frac{1}{6}$, so the correct option is (C).

   We start with the binomial expansion of $\left(2x + \frac{1}{8}\right)^{10}$. The general term (the $(r+1)$-th term) is given by

   $T_{r+1} = \binom{10}{r} (2x)^{10-r} \left(\frac{1}{8}\right)^r.$

   The third term corresponds to $r = 2$ (since $T_3$ means $r+1 = 3$), and the fourth term corresponds to $r = 3$. The problem states these two terms are equal. Our job is to set them equal and solve for $x$.

   ---

   1. Write the third term ($r = 2$)

   $T_3 = \binom{10}{2} (2x)^{10-2} \left(\frac{1}{8}\right)^2 = \binom{10}{2} (2x)^8 \cdot \frac{1}{8^2}.$

   Since $\binom{10}{2} = 45$ and $8^2 = 64$, we have

   $T_3 = 45 \cdot (2x)^8 \cdot \frac{1}{64}.$

   2. Write the fourth term ($r = 3$)

   $T_4 = \binom{10}{3} (2x)^{10-3} \left(\frac{1}{8}\right)^3 = \binom{10}{3} (2x)^7 \cdot \frac{1}{8^3}.$

   Here $\binom{10}{3} = 120$ and $8^3 = 512$, so

   $T_4 = 120 \cdot (2x)^7 \cdot \frac{1}{512}.$

   3. Set them equal

   $45 \cdot (2x)^8 \cdot \frac{1}{64} = 120 \cdot (2x)^7 \cdot \frac{1}{512}.$

   Notice both sides have a factor of $(2x)^7$. We can divide both sides by $(2x)^7$ (provided $x \neq 0$; if $x = 0$ both terms are zero, but that’s not among the options, so it’s safe). This gives

   $45 \cdot (2x) \cdot \frac{1}{64} = 120 \cdot \frac{1}{512}.$

   4. Simplify the equation Multiply both sides by 64 to clear the denominator on the left:

   $45 \cdot (2x) = 120 \cdot \frac{64}{512}.$

   Simplify $\frac{64}{512} = \frac{1}{8}$. So

   $90x = 120 \cdot \frac{1}{8} = 15.$

   5. Solve for $x$

   $x = \frac{15}{90} = \frac{1}{6}.$

   Tip

   A common shortcut: after canceling $(2x)^7$, you’re left with a linear equation in $x$. Always check that $x \neq 0$ before canceling — here it’s fine because zero isn’t an option.

   Watch out

   A classic mistake is to confuse the term number with the index $r$. Remember: the first term has $r = 0$, so the third term has $r = 2$, not $r = 3$. Double-checking this saves errors.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

5. COMEDK 2024Set 2024-A1 markMCQ
   Q.If the sum of the coefficients of the first three terms in the expansion of $\left(x-\frac{a}{x^2}\right)^{12}, x \neq 0$ is 559. Find the value of '$a$' if '$a$' belongs to positive integers (A) 5 (B) 4 (C) $\frac{31}{11}$ (D) 3

   ›Reveal solutionSolution

   The sum of the coefficients of the first three terms in the binomial expansion gives an equation in $a$; solving it yields $a = 3$, so the correct option is (D).

   We are expanding $\left(x - \frac{a}{x^2}\right)^{12}$. The key idea: the "coefficient" of a term in a binomial expansion is the constant factor that multiplies the power of $x$. Here, each term will have the form $\binom{12}{r} x^{12-r} \left(-\frac{a}{x^2}\right)^r$. The coefficient includes both the binomial coefficient and the powers of $a$ and $-1$. The sum of the coefficients of the first three terms means we take $r = 0, 1, 2$ and add their coefficients (ignoring the $x$-part, since we only care about the constant multiplier).

   Let’s work through it step by step.

   1. General term The $(r+1)$-th term in $(x + y)^{12}$ is $\binom{12}{r} x^{12-r} y^r$. Here $y = -\frac{a}{x^2}$. So

   $T_{r+1} = \binom{12}{r} x^{12-r} \left(-\frac{a}{x^2}\right)^r = \binom{12}{r} (-a)^r x^{12-r-2r} = \binom{12}{r} (-a)^r x^{12-3r}.$

   The coefficient of this term (the constant factor in front of the power of $x$) is $\binom{12}{r} (-a)^r$.

   2. First three terms ($r = 0, 1, 2$)

      • For $r = 0$: coefficient = $\binom{12}{0} (-a)^0 = 1$.
      • For $r = 1$: coefficient = $\binom{12}{1} (-a)^1 = 12(-a) = -12a$.
      • For $r = 2$: coefficient = $\binom{12}{2} (-a)^2 = 66 \cdot a^2$ (since $(-a)^2 = a^2$).

   3. Sum of these coefficients

      The problem states this sum is 559. So

   $1 + (-12a) + 66a^2 = 559.$

   Simplify:

   $66a^2 - 12a + 1 = 559 \quad\Rightarrow\quad 66a^2 - 12a - 558 = 0.$

   4. Solve the quadratic Divide through by 6:

   $11a^2 - 2a - 93 = 0.$

   Use the quadratic formula:

   $a = \frac{2 \pm \sqrt{4 + 4 \cdot 11 \cdot 93}}{2 \cdot 11} = \frac{2 \pm \sqrt{4 + 4092}}{22} = \frac{2 \pm \sqrt{4096}}{22}.$

   Since $\sqrt{4096} = 64$, we get

   $a = \frac{2 \pm 64}{22}.$

   So $a = \frac{66}{22} = 3$ or $a = \frac{-62}{22} = -\frac{31}{11}$.

   5. Select the positive integer The problem says $a$ belongs to positive integers. So $a = 3$ is the only valid choice. Option (D).

   Watch out

   A common mistake is forgetting the negative sign from $(-a)^r$ when $r$ is odd. Here for $r=1$ the coefficient is $-12a$, not $+12a$. That would give a different quadratic and a wrong answer.

   Tip

   Notice that the sum of coefficients of the first three terms does not depend on the powers of $x$ — we only need the constant multipliers. This is a standard trick: just set $x = 1$ in each term’s coefficient expression, but here the $x$ exponent doesn’t affect the coefficient itself.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

6. COMEDK 2024Set 2024-M1 markMCQ
   Q.If $\left[{ }^{n+1} C_{r+1}\right]:\left[{ }^n C_r\right]:\left[{ }^{n-1} C_{r-1}\right]=11: 6: 3$ then $n r=$ (A) 30 (B) 40 (C) 20 (D) 50

   ›Reveal solutionSolution

   The key idea is to express the given ratio of binomial coefficients in terms of $n$ and $r$, then solve the resulting system of equations. The product $nr$ is found to be 50.

   We are given:

   $\frac{{}^{n+1}C_{r+1}}{{}^{n}C_r} = \frac{11}{6} \quad \text{and} \quad \frac{{}^{n}C_r}{{}^{n-1}C_{r-1}} = \frac{6}{3} = 2.$

   ---

   Concept and Intuition

   Binomial coefficients have simple factorial forms, and ratios of consecutive ones often collapse into neat linear expressions. Here, the ratio $\frac{{}^{n+1}C_{r+1}}{{}^{n}C_r}$ simplifies to $\frac{n+1}{r+1}$, and $\frac{{}^{n}C_r}{{}^{n-1}C_{r-1}}$ simplifies to $\frac{n}{r}$. This turns a combinatorial ratio into a pair of algebraic equations, which we can solve directly for $n$ and $r$.

   ---

   Step-by-step solution

   1. Write the first ratio

   $\frac{{}^{n+1}C_{r+1}}{{}^{n}C_r} = \frac{11}{6}$

   Using the factorial definition:

   ${}^{n+1}C_{r+1} = \frac{(n+1)!}{(r+1)!(n-r)!}, \quad {}^{n}C_r = \frac{n!}{r!(n-r)!}$

   Their ratio simplifies:

   $\frac{(n+1)!}{(r+1)!(n-r)!} \cdot \frac{r!(n-r)!}{n!} = \frac{n+1}{r+1}$

   So we have:

   $\frac{n+1}{r+1} = \frac{11}{6} \quad \Rightarrow \quad 6(n+1) = 11(r+1) \quad \Rightarrow \quad 6n + 6 = 11r + 11$

   $6n - 11r = 5 \qquad(1)$

   2. Write the second ratio

   $\frac{{}^{n}C_r}{{}^{n-1}C_{r-1}} = \frac{6}{3} = 2$

   Simplify similarly:

   ${}^{n}C_r = \frac{n!}{r!(n-r)!}, \quad {}^{n-1}C_{r-1} = \frac{(n-1)!}{(r-1)!(n-r)!}$

   Ratio:

   $\frac{n!}{r!(n-r)!} \cdot \frac{(r-1)!(n-r)!}{(n-1)!} = \frac{n}{r}$

   Hence:

   $\frac{n}{r} = 2 \quad \Rightarrow \quad n = 2r \qquad(2)$

   3. Solve the system Substitute $n = 2r$ into equation (1):

   $6(2r) - 11r = 5 \quad \Rightarrow \quad 12r - 11r = 5 \quad \Rightarrow \quad r = 5$

   Then $n = 2r = 10$.

   4. Compute $nr$

   $nr = 10 \times 5 = 50$

   ---

   Tip

   The ratio $\frac{{}^{n}C_r}{{}^{n-1}C_{r-1}} = \frac{n}{r}$ is a handy shortcut — it comes from canceling almost everything in the factorial expansion.

   Watch out

   A common mistake is to misalign the indices: ensure you correctly identify which coefficient corresponds to which term in the ratio. Here the middle term is ${}^{n}C_r$, so the second ratio is exactly $\frac{{}^{n}C_r}{{}^{n-1}C_{r-1}}$.

   ---

   ✓Final answer

   The correct option is (D).

   ANSWER: D

7. COMEDK 2024Set 2024-M1 markMCQ
   Q.The coefficient of the third term in the expansion of $\left(x^2-\frac{1}{4}\right)^n$, when expanded in the descending power of $x$ is 31, then $n$ is (A) 31 (B) 16 (C) 30 (D) 32

   ›Reveal solutionSolution

   The third term's coefficient is $\binom{n}{2}\cdot\tfrac{1}{16}=31$, giving $\binom{n}{2}=496$ and $n=32$ (option D).

   In the descending-power expansion of $\left(x^2-\dfrac{1}{4}\right)^n$, the general term is

   $T_{r+1}=\binom{n}{r}\left(x^2\right)^{n-r}\left(-\frac{1}{4}\right)^{r}$

   The third term corresponds to $r=2$:

   $T_3=\binom{n}{2}\left(x^2\right)^{n-2}\left(-\frac{1}{4}\right)^{2}=\binom{n}{2}\cdot\frac{1}{16}\,x^{2(n-2)}$

   Its coefficient is given as $31$:

   $\frac{1}{16}\binom{n}{2}=31\quad\Longrightarrow\quad \binom{n}{2}=496$

   $\frac{n(n-1)}{2}=496\;\Longrightarrow\; n(n-1)=992=32\times 31\;\Longrightarrow\; n=32$

   ✓Final answer

   $n=32$ — option (D).

8. COMEDK 2023Set 2023-M1 markMCQ
   Q.

   [!FORMULA] $2^{3 n}-7 n-1 \text { is divisible by }$

   (A) 64 (B) 36 (C) 49 (D) 25

   ›Reveal solutionSolution

   $2^{3n}-7n-1=8^n-7n-1$ is always divisible by $49$ (binomial expansion of $(1+7)^n$).

   Write $2^{3n}=(2^3)^n=8^n=(1+7)^n$. By the binomial theorem:

   $8^n=1+\binom{n}{1}7+\binom{n}{2}7^2+\binom{n}{3}7^3+\cdots$

   So

   $8^n-7n-1=\binom{n}{2}7^2+\binom{n}{3}7^3+\cdots$

   Every remaining term contains $7^2=49$ as a factor, hence the whole expression is divisible by $49$.

   Check: $n=2\Rightarrow 64-14-1=49$; $n=3\Rightarrow 512-21-1=490=49\cdot10$. Both divisible by $49$.

   ✓Final answer

   The correct option is (C) — 49

9. COMEDK 2023Set 2023-M1 markMCQ
   Q.The total number of terms in the expansion of $(x+y)^{60}+(x-y)^{60}$ is (A) 60 (B) 61 (C) 30 (D) 31

   ›Reveal solutionSolution

   Adding the two expansions cancels all odd-power (in $y$) terms and doubles the even ones, leaving terms for $k=0,2,\dots,60$ — that is $31$ terms.

   $(x+y)^{60}=\sum_{k=0}^{60}\binom{60}{k}x^{60-k}y^{k}$ and $(x-y)^{60}=\sum_{k=0}^{60}\binom{60}{k}x^{60-k}(-y)^{k}$.

   Adding: terms with odd $k$ have opposite signs and cancel; terms with even $k$ add. The surviving exponents are $k=0,2,4,\dots,60$.

   Number of even values from $0$ to $60$: $\frac{60}{2}+1=31$.

   ✓Final answer

   The correct option is (D) — 31

10. COMEDK 2023Set 2023-M1 markMCQ
    Q.The coefficient of $x^{29}$ in the expansion of $\left(1-3 x+3 x^2-x^3\right)^{15}$ is (A) ${ }^{45} C_{29}$ (B) ${ }^{45} C_{28}$ (C) $-{ }^{45} C_{16}$ (D) ${ }^{45} C_{30}$

    ›Reveal solutionSolution

    Recognise $1-3x+3x^2-x^3=(1-x)^3$, so the whole thing is $(1-x)^{45}$; the $x^{29}$ coefficient is $-\binom{45}{16}$.

    Note $1-3x+3x^2-x^3=(1-x)^3$. Therefore:

    $\left(1-3x+3x^2-x^3\right)^{15}=\big((1-x)^3\big)^{15}=(1-x)^{45}.$

    General term: $\binom{45}{k}(-x)^{k}=\binom{45}{k}(-1)^k x^k$. For $x^{29}$ take $k=29$:

    $\binom{45}{29}(-1)^{29}=-\binom{45}{29}=-\binom{45}{16}$

    using $\binom{45}{29}=\binom{45}{16}$.

    ✓Final answer

    The correct option is (C) — $-{ }^{45} C_{16}$

11. COMEDK 2023Set 2023-M1 markMCQ
    Q.In the expansion of $\left(1+3 x+3 x^2+x^3\right)^{2 n}$, the term which has greatest binomial coefficient, is (A) $(3 n)$ th term (B) $(3 n+1)$ th term (C) $(3 n-1)$ th term (D) $(3 n+2)$ th term

    ›Reveal solutionSolution

    The base collapses to $(1+x)^{6n}$; its central (largest) binomial coefficient $\binom{6n}{3n}$ is the $(3n+1)$th term.

    Since $1+3x+3x^2+x^3=(1+x)^3$:

    $\left(1+3x+3x^2+x^3\right)^{2n}=\big((1+x)^3\big)^{2n}=(1+x)^{6n}.$

    The expansion of $(1+x)^{6n}$ has $6n+1$ terms, with binomial coefficients $\binom{6n}{k}$, $k=0,\dots,6n$. These are maximal at the central value $k=3n$.

    The term with $k=3n$ is the $(k+1)=(3n+1)$th term.

    ✓Final answer

    The correct option is (B) — $(3 n+1)$ th term

12. COMEDK 2022Set 20221 markMCQ
    Q.The total number of terms in the expansion of ${(x + y)^{100}} + {(x - y)^{100}}$ is (A) 49 (B) 50 (C) 51 (D) 99

    ›Reveal solutionSolution

    Number of such terms = 100/2 + 1 = 51.

    Concept: In (x + y)ⁿ + (x − y)ⁿ the terms with odd powers of y cancel; the even ones double.

    For n = 100 the surviving terms are those with y⁰, y², y⁴, …, y¹⁰⁰ — i.e. exponents 0, 2, 4, …, 100.

    Number of such terms = 100/2 + 1 = 51.

    ✓Final answer

    The correct option is (C) — 51

    ANSWER: C