Q.Evaluate $\displaystyle\sum_{k=1}^{11} (2 + 3^k)$.

Showing the 12 most recent of 21 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.$1.5 + 2.8 + 3.11 + \cdots$ to $n$ terms $=$ (A) $\dfrac{n(n+1)(2n+3)}{2}$ (B) $\dfrac{n(n+1)(2n+1)}{2}$ (C) $\dfrac{n(n+1)(n+2)}{3}$ (D) $\dfrac{n(n+1)(n+3)}{3}$

   ›Reveal solutionSolution

   Each term is (position)$\times$(an AP with first term 5, common difference 3); summing the resulting quadratic $3k^2+2k$ gives $\frac{n(n+1)(2n+3)}{2}$. Answer: (A).

   Concept and Intuition

   The series is $1\cdot 5 + 2\cdot 8 + 3\cdot 11 + \cdots$. The first factor in each term is just the position $k=1,2,3,\dots$. The second factor is $5, 8, 11, \dots$, an arithmetic progression with first term $5$ and common difference $3$, so its $k$-th term is $5+3(k-1) = 3k+2$. So the general term of the series is $T_k = k(3k+2)$.

   To sum $n$ such terms, expand into a polynomial in $k$ and use the standard sums $\sum k = \frac{n(n+1)}{2}$ and $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$.

   Step-by-Step Solution

   1. General term: $T_k = k \cdot (3k+2) = 3k^2 + 2k$.
   2. Sum: $S_n = \sum_{k=1}^n (3k^2+2k) = 3\sum_{k=1}^n k^2 + 2\sum_{k=1}^n k$.
   3. $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$, so $3\sum k^2 = \frac{n(n+1)(2n+1)}{2}$.
   4. $\sum k = \frac{n(n+1)}{2}$, so $2\sum k = n(n+1)$. …

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2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If $\dfrac{3}{4} + \dfrac{15}{16} + \dfrac{63}{64} + \cdots n \text{ terms} = \dfrac{939}{256}$, then $5n =$ (A) $55$ (B) $20$ (C) $15$ (D) $35$

   ›Reveal solutionSolution

   The series is a sum of fractions of the form $\frac{2^{2k} - 1}{2^{2k}} = 1 - \frac{1}{4^k}$. Summing $n$ such terms gives $n - \frac{1}{3}\left(1 - \frac{1}{4^n}\right)$. Setting this equal to $\frac{939}{256}$ yields $n = 4$, so $5n = 20$.

   We are given:

   $\frac{3}{4} + \frac{15}{16} + \frac{63}{64} + \cdots \text{ to } n \text{ terms} = \frac{939}{256}$

   and asked for $5n$.

   ---

   Concept and Intuition

   Each term looks like a fraction close to 1. Notice:

   • $\frac{3}{4} = 1 - \frac{1}{4}$
   • $\frac{15}{16} = 1 - \frac{1}{16}$
   • $\frac{63}{64} = 1 - \frac{1}{64}$

   The denominators are powers of 4: $4^1, 4^2, 4^3, \dots$. So the $k$-th term is:

   $\frac{4^k - 1}{4^k} = 1 - \frac{1}{4^k}$

   Thus the sum of $n$ terms is:

   $S_n = \sum_{k=1}^n \left(1 - \frac{1}{4^k}\right) = n - \sum_{k=1}^n \frac{1}{4^k}$

   The second sum is a finite geometric series. This transforms the problem into simple algebra.

   ---

   Step-by-step solution

   1. Write the general term

      The numerators are $3, 15, 63, \dots$ which are $4-1, 16-1, 64-1, \dots$ i.e. $4^k - 1$ for $k = 1,2,3,\dots$.

      So term $T_k = \dfrac{4^k - 1}{4^k} = 1 - \dfrac{1}{4^k}$.

   2. Sum the series

   $S_n = \sum_{k=1}^n \left(1 - \frac{1}{4^k}\right) = n - \sum_{k=1}^n \frac{1}{4^k}$

   3. Sum the geometric series $\sum_{k=1}^n \frac{1}{4^k}$ is a geometric series with first term $a = \frac{1}{4}$, ratio $r = \frac{1}{4}$, and $n$ terms.

   $\sum_{k=1}^n \frac{1}{4^k} = \frac{\frac{1}{4}\left(1 - \frac{1}{4^n}\right)}{1 - \frac{1}{4}} = \frac{\frac{1}{4}\left(1 - \frac{1}{4^n}\right)}{\frac{3}{4}} = \frac{1}{3}\left(1 - \frac{1}{4^n}\right)$

   4. Write the total sum

   $S_n = n - \frac{1}{3}\left(1 - \frac{1}{4^n}\right)$

   5. Set equal to given value

   $n - \frac{1}{3}\left(1 - \frac{1}{4^n}\right) = \frac{939}{256}$

   6. Solve for $n$ Multiply by 3:

   $3n - \left(1 - \frac{1}{4^n}\right) = \frac{2817}{256}$

   $3n - 1 + \frac{1}{4^n} = \frac{2817}{256}$

   $3n + \frac{1}{4^n} = \frac{2817}{256} + 1 = \frac{3073}{256}$

   Now test small integer $n$: …

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3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The sum $S_n$ of the $n$ terms of the series $1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + 2 \times 6^2 + \cdots$ is given by (A) $S_n = \dfrac{n^2(n+1)^2}{2}$, $n$ is even ; $S_n = \dfrac{n(n+1)}{2}$, $n$ is odd (B) $S_n = \dfrac{n(n^2+1)}{2}$, $n$ is even ; $S_n = \dfrac{n^2(n+1)}{2}$, $n$ is odd (C) $S_n = \dfrac{n(n+1)^2}{2}$, $n$ is even ; $S_n = \dfrac{n^2(n+1)}{2}$, $n$ is odd (D) $S_n = \dfrac{n^2(n^2+1)}{2}$, $n$ is even ; $S_n = \dfrac{n(n+1)^2}{2}$, $n$ is odd

   ›Reveal solutionSolution

   The series alternates between a square and twice the next square; by grouping pairs and handling the odd‑term case separately, we derive closed forms that match option (C).

   ---

   Concept and Intuition

   The series is:

   $1^2 + 2\cdot 2^2 + 3^2 + 2\cdot 4^2 + 5^2 + 2\cdot 6^2 + \cdots$

   Notice the pattern:

   • Terms at odd positions (1st, 3rd, 5th, …) are just squares: $1^2, 3^2, 5^2, \dots$
   • Terms at even positions (2nd, 4th, 6th, …) are twice the square of the even number: $2\cdot 2^2, 2\cdot 4^2, 2\cdot 6^2, \dots$

   So the series is not a simple sum of squares — it’s a hybrid. The natural approach is to group terms in pairs (odd + even) because each pair has a neat structure. Then we handle the case when $n$ is odd separately (since the last term will be an unpaired odd‑position square).

   ---

   Step‑by‑Step Derivation

   1. Write the general term

   Let the $k$-th pair be $(2k-1)^2$ (odd) and $2\cdot (2k)^2$ (even).

   So the $k$-th pair sum is:

   $P_k = (2k-1)^2 + 2\cdot (2k)^2$

   2. Simplify the pair sum

   $\begin{aligned} P_k &= (4k^2 - 4k + 1) + 2\cdot (4k^2) \\ &= 4k^2 - 4k + 1 + 8k^2 \\ &= 12k^2 - 4k + 1 \end{aligned}$

   3. Sum for $n$ even

   If $n$ is even, say $n = 2m$, then we have exactly $m$ complete pairs. So:

   $S_{2m} = \sum_{k=1}^{m} (12k^2 - 4k + 1)$

   Use standard formulas:

   $\sum_{k=1}^m k^2 = \frac{m(m+1)(2m+1)}{6}, \quad \sum_{k=1}^m k = \frac{m(m+1)}{2}, \quad \sum_{k=1}^m 1 = m$

   Thus:

   $\begin{aligned} S_{2m} &= 12\cdot\frac{m(m+1)(2m+1)}{6} - 4\cdot\frac{m(m+1)}{2} + m \\ &= 2m(m+1)(2m+1) - 2m(m+1) + m \end{aligned}$

   Factor $m$ out:

   $S_{2m} = m\bigl[2(m+1)(2m+1) - 2(m+1) + 1\bigr]$

   Simplify inside:

   $\begin{aligned} 2(m+1)(2m+1) - 2(m+1) + 1 &= 2(m+1)(2m+1 - 1) + 1 \\ &= 2(m+1)(2m) + 1 \\ &= 4m(m+1) + 1 \end{aligned}$

   So:

   $S_{2m} = m\bigl(4m(m+1) + 1\bigr) = 4m^2(m+1) + m$

   Now replace $m = n/2$:

   $S_n = 4\left(\frac{n}{2}\right)^2\left(\frac{n}{2}+1\right) + \frac{n}{2} = 4\cdot\frac{n^2}{4}\cdot\frac{n+2}{2} + \frac{n}{2} = \frac{n^2(n+2)}{2} + \frac{n}{2} = \frac{n}{2}\bigl(n(n+2) + 1\bigr)$

   Simplify:

   $n(n+2) + 1 = n^2 + 2n + 1 = (n+1)^2$

   Hence:

   $\boxed{S_n = \frac{n(n+1)^2}{2} \quad \text{for } n \text{ even}}$

   ---

   4. Sum for $n$ odd

   If $n$ is odd, say $n = 2m+1$, then we have $m$ complete pairs plus one extra term: the $(2m+1)$-th term, which is an odd‑position square: $(2m+1)^2$.

   So:

   $S_{2m+1} = S_{2m} + (2m+1)^2$

   We already have $S_{2m} = 4m^2(m+1) + m$. Add $(2m+1)^2 = 4m^2 + 4m + 1$:

   $$$$

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4. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.$\lim\limits_{n\to\infty} \dfrac{1+2^4+3^4+\cdots+n^4}{n^5} - \lim\limits_{n\to\infty} \dfrac{1+2^3+3^3+\cdots+n^3}{n^5} =$ (A) $\dfrac{1}{5}$ (B) $\dfrac{1}{4}$ (C) $\dfrac{1}{20}$ (D) $0$

   ›Reveal solutionSolution

   Both limits are Riemann-sum-style power-sum limits; the $k^4$ sum grows like $n^5/5$ giving $1/5$, while the $k^3$ sum only grows like $n^4$, so dividing by $n^5$ sends the second limit to $0$ — the answer is $1/5$.

   Concept and Intuition

   A sum $\sum_{k=1}^n k^p$ grows like $\dfrac{n^{p+1}}{p+1}$ for large $n$ (this is exactly the Riemann-sum statement $\frac{1}{n^{p+1}}\sum k^p\to\int_0^1 x^p\,dx=\frac{1}{p+1}$). Dividing by $n^5$ picks out only the $p=4$ sum as contributing a finite nonzero limit; the $p=3$ sum, being one power of $n$ smaller, vanishes in the same limit.

   Step-by-Step Solution

   1. Recognize $\displaystyle\lim_{n\to\infty}\frac{1^4+2^4+\cdots+n^4}{n^5}=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\left(\frac{k}{n}\right)^4=\int_0^1x^4dx=\frac15$.
   2. Recognize $\displaystyle\lim_{n\to\infty}\frac{1^3+2^3+\cdots+n^3}{n^5}=\lim_{n\to\infty}\frac{1}{n^2}\cdot\frac{1}{n}\sum_{k=1}^n\left(\frac{k}{n}\right)^3=\lim_{n\to\infty}\frac1{n^2}\int_0^1x^3dx=\lim_{n\to\infty}\frac{1}{4n^2}=0$. …

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5. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.For all $n \in N$, $\dfrac{1}{a(a+d)} + \dfrac{1}{(a+d)(a+2d)} + \dfrac{1}{(a+2d)(a+3d)} + \cdots$ upto $n$ terms $=$ (A) $\dfrac{nd}{a+d}$ (B) $\dfrac{nd}{a(a+nd)}$ (C) $\dfrac{n}{a(a+nd)}$ (D) $\dfrac{d}{a+nd}$

   ›Reveal solutionSolution

   The sum telescopes when each term is rewritten as a difference of two simpler fractions. The final result is $\dfrac{n}{a(a+nd)}$, which corresponds to option (C).

   ---

   Concept and Intuition (Sum of Series)

   When you see a sum of fractions whose denominators are products of terms in arithmetic progression — like $a(a+d)$, $(a+d)(a+2d)$, etc. — your first instinct should be telescoping.

   The trick: each term can be split into a difference of two fractions whose denominators are the two factors. Then, when you add them up, most terms cancel, leaving only the first and last.

   Here, the general term is $\dfrac{1}{(a+(k-1)d)(a+kd)}$. Notice that:

   $\frac{1}{(a+(k-1)d)(a+kd)} = \frac{1}{d}\left( \frac{1}{a+(k-1)d} - \frac{1}{a+kd} \right)$

   Why? Because if you combine the right-hand side over a common denominator, you get exactly the left-hand side. This is the classic partial-fraction decomposition for such products.

   ---

   Step-by-step solution

   1. Write the sum explicitly The series is:

   $S_n = \frac{1}{a(a+d)} + \frac{1}{(a+d)(a+2d)} + \frac{1}{(a+2d)(a+3d)} + \cdots + \frac{1}{(a+(n-1)d)(a+nd)}$

   There are $n$ terms, and the last denominator uses $a+(n-1)d$ and $a+nd$.

   2. Decompose each term For any integer $k$ from $1$ to $n$, the $k$-th term is:

   $\frac{1}{(a+(k-1)d)(a+kd)} = \frac{1}{d}\left( \frac{1}{a+(k-1)d} - \frac{1}{a+kd} \right)$

   This is verified by cross-multiplying:

   $\frac{1}{d}\left( \frac{a+kd - (a+(k-1)d)}{(a+(k-1)d)(a+kd)} \right) = \frac{1}{d} \cdot \frac{d}{(a+(k-1)d)(a+kd)} = \frac{1}{(a+(k-1)d)(a+kd)}$

   3. Write the sum using the decomposition

   $S_n = \frac{1}{d} \left[ \left( \frac{1}{a} - \frac{1}{a+d} \right) + \left( \frac{1}{a+d} - \frac{1}{a+2d} \right) + \cdots + \left( \frac{1}{a+(n-1)d} - \frac{1}{a+nd} \right) \right]$

   4. Observe the telescoping cancellation …

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6. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.$-1 + \dfrac{7}{10}\cdot 2^2 - \dfrac{7\cdot 9}{10\cdot 15}\cdot 2^3 + \dfrac{7\cdot 9\cdot 11}{10\cdot 15\cdot 20}\cdot 2^4 - \cdots \infty =$ (A) $\left(\dfrac{25}{81}\right)^{1/5}$ (B) $\dfrac{25\sqrt{5}}{243}$ (C) $\left(\dfrac{81}{25}\right)^{1/5}$ (D) $\dfrac{243}{25\sqrt{5}}$

   ›Reveal solutionSolution

   The alternating series with those specific factorial-style numerators/denominators and increasing powers of 2 is a generalised binomial series in disguise; summing it in closed form gives $(5/9)^{5/2}=25\sqrt5/243$.

   Concept and Intuition

   Series whose terms have numerators like $7\cdot9\cdot11\cdots$ and denominators like $10\cdot15\cdot20\cdots$ are built from the generalised binomial coefficients that appear in expansions such as $(1-x)^{-n} = \sum_k \dfrac{n(n+1)(n+2)\cdots(n+k-1)}{k!}x^k$. Recognising the pattern (numerator steps of 2, denominator steps of 5) points to an underlying exponent of $n=7/2$, and the extra $\frac{1}{(k+1)!}$-type factor (compared to a plain binomial term) signals that this is the integral of a $(1-x)^{-7/2}$ series, which integrates cleanly to a $(1-x)^{-5/2}$ expression.

   Step-by-Step Solution

   1. Write the general term (for $k\ge1$) of the series (ignoring sign) as $\dfrac{\prod_{i=0}^{k-1}(7+2i)}{\prod_{i=0}^{k-1}(10+5i)}\cdot 2^{k+1}$.
   2. Factor out constants: $\prod(7+2i) = 2^k\left(\tfrac72\right)_k$ and $\prod(10+5i)=5^k(k+1)!$, where $\left(\tfrac72\right)_k$ is the rising factorial from $7/2$.
   3. This shows the series is $-2\sum_{k\ge0} y^k \dfrac{(7/2)_k}{(k+1)!}$ for a suitable $y=-4/5$, i.e. essentially the term-by-term integral of the binomial series for $(1-x)^{-7/2}$. …

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7. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.If $(1+x)^n = \displaystyle\sum_{r=0}^{n} C_r x^r$, then the value of $C_0 + (C_0+C_1) + (C_0+C_1+C_2) + \ldots + (C_0+C_1+C_2+\ldots+C_n)$ is (A) $n2^{n-1}$ (B) $2^n + n$ (C) $(n+2)2^n$ (D) $(n+2)2^{n-1}$

   ›Reveal solutionSolution

   Swapping the order of a double summation turns the nested partial-sum expression into a single sum weighted by $(n-r+1)$, which collapses using the two standard binomial identities $\sum C_r=2^n$ and $\sum rC_r=n2^{n-1}$ to give $(n+2)2^{n-1}$.

   Concept and Intuition

   The expression $C_0+(C_0+C_1)+(C_0+C_1+C_2)+\ldots+(C_0+\ldots+C_n)$ is a sum of partial sums of the binomial coefficients. Rather than summing partial sums directly, it's far cleaner to ask: 'how many times does each individual $C_r$ get counted across all these partial sums?' — this reframes a nested sum as a single weighted sum, which then falls to two well-known binomial identities.

   Step-by-Step Solution

   1. Define $S_k=\sum_{r=0}^{k}C_r$ (the $k$-th partial sum). We want $S=\sum_{k=0}^{n}S_k$.
   2. Swap the order of summation: $S=\sum_{k=0}^{n}\sum_{r=0}^{k}C_r=\sum_{r=0}^{n}C_r\cdot\#\{k : r\le k\le n\}$.
   3. For fixed $r$, the number of valid $k$ (with $k$ ranging from $r$ to $n$) is $(n-r+1)$. So $S=\sum_{r=0}^{n}(n-r+1)\,C_r$.
   4. Split this: $S=(n+1)\sum_{r=0}^n C_r-\sum_{r=0}^n r\,C_r$.
   5. Use the standard identities: $\sum_{r=0}^n C_r=2^n$ (sum of a row of Pascal's triangle) and $\sum_{r=0}^n r\,C_r=n\cdot2^{n-1}$ (a classic identity, provable by differentiating $(1+x)^n$ and setting $x=1$).
   6. Substitute: $S=(n+1)2^n-n\cdot2^{n-1}=2^{n-1}\left[2(n+1)-n\right]=2^{n-1}(2n+2-n)=2^{n-1}(n+2)$. …

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8. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.$\displaystyle\sum_{k=1}^{n} k(k+1)(k+2)\ldots(k+r-1) =$ (A) $\dfrac{n(n+1)(n+2)\ldots(n+r)}{r+1}$ (B) $\dfrac{n(n+1)(n+2)\ldots(n+r-1)}{r}$ (C) $\dfrac{n(n+1)(n+2)\ldots(n+r+1)}{r+1}$ (D) $\dfrac{n(n+1)(n+2)\ldots 2n}{2n+1}$

   ›Reveal solutionSolution

   Summing $k(k+1)\cdots(k+r-1)$ from $1$ to $n$ telescopes to $\dfrac{n(n+1)\cdots(n+r)}{r+1}$ — option (A).

   Concept and Intuition

   A product of $r$ consecutive integers starting at $k$ behaves like a discrete analogue of $x^r$, and just as $\int x^r\,dx=\dfrac{x^{r+1}}{r+1}$, there is a discrete 'antidifference' for such falling-factorial-type products: $k(k+1)\cdots(k+r-1)$ is (up to the constant $\dfrac1{r+1}$) the first difference of $k(k+1)\cdots(k+r)$. This lets the whole sum collapse (telescope) instead of needing to expand and sum power series term by term.

   Step-by-Step Solution

   1. Define $T_k = k(k+1)(k+2)\cdots(k+r-1)$, the general term.
   2. Consider $U_k = k(k+1)(k+2)\cdots(k+r)$ (one factor more, up to $k+r$) and $U_{k-1} = (k-1)k(k+1)\cdots(k+r-1)$.
   3. Compute $U_k - U_{k-1} = k(k+1)\cdots(k+r-1)\big[(k+r) - (k-1)\big] = k(k+1)\cdots(k+r-1)\cdot(r+1) = (r+1)T_k$.
   4. So $T_k = \dfrac{U_k - U_{k-1}}{r+1}$, and summing from $k=1$ to $n$ telescopes: $\sum_{k=1}^n T_k = \dfrac{U_n - U_0}{r+1}$.
   5. $U_0 = 0\cdot1\cdots r = 0$ (it contains the factor $0$), so $\sum_{k=1}^n T_k = \dfrac{U_n}{r+1} = \dfrac{n(n+1)(n+2)\cdots(n+r)}{r+1}$. …

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9. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.$1+\dfrac{4}{15}+\dfrac{4.10}{15.30}+\dfrac{4.10.16}{15.30.45}+\ldots \infty=$ (A) $\left(\dfrac{3}{5}\right)^{2/3}$ (B) $\left(\dfrac{5}{3}\right)^{2/3}$ (C) $\left(\dfrac{3}{5}\right)^{3/2}$ (D) $\left(\dfrac{5}{3}\right)^{3/2}$

   ›Reveal solutionSolution

   This tests recognizing a series as a generalized binomial expansion $(1-x)^{-p}$ by matching the ratio of consecutive terms. Answer: $\left(\frac{5}{3}\right)^{2/3}$.

   Concept and Intuition

   The generalized binomial theorem gives $(1-x)^{-p} = 1 + px + \dfrac{p(p+1)}{2!}x^2 + \dfrac{p(p+1)(p+2)}{3!}x^3+\cdots$, whose $n$-th to $(n-1)$-th term ratio is $\dfrac{p+n-1}{n}x$. Many exam series with steadily-incrementing numerator/denominator factors are exactly this expansion in disguise; the trick is to compute the term ratio and match it to $\frac{p+n-1}{n}x$.

   Step-by-Step Solution

   1. Terms: $t_1=\frac{4}{15}$, $t_2=\frac{4\cdot10}{15\cdot30}$, $t_3=\frac{4\cdot10\cdot16}{15\cdot30\cdot45}$. The numerator factors increase by 6 each time ($4,10,16,\ldots$) and the denominator factors are multiples of 15 ($15,30,45,\ldots$).
   2. The ratio of the $n$-th term to the $(n-1)$-th term is $\dfrac{4+6(n-1)}{15n}=\dfrac{6n-2}{15n}$. …

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10. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.$\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+\ldots$ to 24 terms $=$ (A) $\dfrac{23}{147}$ (B) $\dfrac{6}{35}$ (C) $\dfrac{6}{37}$ (D) $\dfrac{8}{51}$

    ›Reveal solutionSolution

    This is a telescoping series; splitting each term by partial fractions collapses the 24-term sum to $\dfrac{8}{51}$.

    Concept and Intuition

    Terms of the form $\dfrac{1}{(2n+1)(2n+3)}$ split into a difference of two simpler fractions differing by a fixed "step" (here, from $\tfrac{1}{2n+1}$ to $\tfrac{1}{2n+3}$). When you sum consecutive terms, the middle values cancel in pairs (telescoping), leaving only the very first and very last pieces.

    Step-by-Step Solution

    1. Write the series as $\displaystyle\sum_{n=1}^{24}\frac{1}{(2n+1)(2n+3)}$, since the terms are $\tfrac{1}{3\cdot5},\tfrac{1}{5\cdot7},\tfrac{1}{7\cdot9},\ldots$ (i.e. $n=1,2,3,\ldots$).
    2. Partial fractions: $\dfrac{1}{(2n+1)(2n+3)} = \dfrac12\left(\dfrac{1}{2n+1}-\dfrac{1}{2n+3}\right)$.
    3. Sum telescopes: $\frac12\left[\left(\frac13-\frac15\right)+\left(\frac15-\frac17\right)+\cdots+\left(\frac{1}{49}-\frac{1}{51}\right)\right] = \frac12\left(\frac13-\frac{1}{51}\right)$ …

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11. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
    Q.If $S_n = 1^3+2^3+\ldots+n^3$ and $T_n = 1+2+\ldots+n$, then (A) $S_n = T_{n^3}$ (B) $S_n = T_n^3$ (C) $S_n = T_{n^2}$ (D) $S_n = T_n^2$

    ›Reveal solutionSolution

    This is the standard identity that the sum of the first $n$ cubes equals the square of the sum of the first $n$ natural numbers: $S_n=T_n^2$.

    Concept and Intuition

    $T_n = 1+2+\cdots+n = \dfrac{n(n+1)}{2}$ is the familiar sum of the first $n$ natural numbers. There is a classical, easily-verified identity that the sum of the first $n$ cubes equals the square of this same sum: $1^3+2^3+\cdots+n^3 = \left(1+2+\cdots+n\right)^2$. This can be seen directly from the standard closed form $S_n = \left(\dfrac{n(n+1)}{2}\right)^2 = T_n^2$.

    Step-by-Step Solution

    1. Recall the standard formula: $S_n = \displaystyle\sum_{k=1}^n k^3 = \left(\dfrac{n(n+1)}{2}\right)^2$.
    2. Recall $T_n = \displaystyle\sum_{k=1}^n k = \dfrac{n(n+1)}{2}$.
    3. Substitute: $S_n = \left(T_n\right)^2 = T_n^2$.
    4. Quick numeric check with $n=3$: $T_3=1+2+3=6$, $T_3^2=36$; and $S_3=1^3+2^3+3^3=1+8+27=36$. Matches, confirming $S_n=T_n^2$. …

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12. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
    Q.$1 - \dfrac{2}{3} + \dfrac{2.4}{3.6} - \dfrac{2.4.6}{3.6.9} + \cdots \infty =$ (A) $\dfrac{3}{5}$ (B) $\left(\dfrac{2}{5}\right)^{2/3}$ (C) $\dfrac{2}{5}$ (D) $\left(\dfrac{3}{5}\right)^{2/3}$

    ›Reveal solutionSolution

    This series looks like a generalized binomial expansion but each term actually collapses to a plain geometric ratio; the sum is $\dfrac{3}{5}$.

    Concept and Intuition

    Numerator of the $n$-th term is $2\cdot4\cdot6\cdots(2n) = 2^n n!$ and the denominator is $3\cdot6\cdot9\cdots(3n) = 3^n n!$. The $n!$ cancels, leaving exactly $\left(\dfrac23\right)^n$ — a clean geometric progression, not a fractional-power binomial series as the answer choices might suggest.

    Step-by-Step Solution

    1. Term $0$: $1$. Term $1$: $\dfrac{2}{3}$. Term $2$: $\dfrac{2\cdot4}{3\cdot6}=\dfrac{8}{18}=\dfrac{4}{9}=\left(\dfrac23\right)^2$. Term $3$: $\dfrac{2\cdot4\cdot6}{3\cdot6\cdot9}=\dfrac{48}{162}=\dfrac{8}{27}=\left(\dfrac23\right)^3$.
    2. So the series is $\sum_{n=0}^\infty (-1)^n\left(\dfrac23\right)^n$, a geometric series with first term $a=1$ and common ratio $r=-\dfrac23$.
    3. Since $|r|<1$, it converges to $S=\dfrac{a}{1-r} = \dfrac{1}{1-(-2/3)} = \dfrac{1}{5/3}=\dfrac35$. …

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