Method of Differences & Telescoping Series — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics Method of Differences & Telescoping Series MCQs with step-by-step solutions (20 questions). Part of Sequences and Series. Practise online on Prepizo — no login needed.
▶ Practise Method of Differences & Telescoping Series online (free)
Questions with solutions
Q1 — Method of Differences & Telescoping Series · medium · theory
The partial-fraction decomposition of $\dfrac{1}{r(r+1)}$ is:
A. $\dfrac{1}{r} - \dfrac{1}{r-1}$
B. $\dfrac1r - \dfrac{1}{r+1}$ ✓ Correct
C. $\dfrac{1}{r+1} - \dfrac1r$
D. $\dfrac1r + \dfrac{1}{r+1}$
Solution: $\dfrac{1}{r(r+1)} = \dfrac1r - \dfrac{1}{r+1}$, the key to telescoping this series.
Q2 — Method of Differences & Telescoping Series · medium · numerical
The sum $\dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \cdots + \dfrac{1}{9 \cdot 10}$ is:
A. $\dfrac{10}{11}$
B. $\dfrac{9}{10}$ ✓ Correct
C. $\dfrac{1}{10}$
D. $1$
Solution: Telescoping: $1 - \dfrac{1}{10} = \dfrac{9}{10}$.
Q3 — Method of Differences & Telescoping Series · medium · numerical
The sum to infinity $\sum_{r=1}^{\infty} \dfrac{1}{r(r+1)}$ is:
A. $\dfrac12$
B. $1$ ✓ Correct
C. $\dfrac34$
D. $2$
Solution: The partial sum is $1 - \dfrac{1}{n+1} \to 1$ as $n \to \infty$.
Q4 — Method of Differences & Telescoping Series · hard · numerical
The sum $\sum_{r=1}^{n} \dfrac{1}{(2r-1)(2r+1)}$ equals:
A. $\dfrac{2n}{2n+1}$
B. $\dfrac{n}{2n-1}$
C. $\dfrac{n}{2n+1}$ ✓ Correct
D. $\dfrac{1}{2n+1}$
Solution: $\dfrac{1}{(2r-1)(2r+1)} = \tfrac12\left(\dfrac{1}{2r-1} - \dfrac{1}{2r+1}\right)$; telescoping gives $\dfrac12\left(1 - \dfrac{1}{2n+1}\right) = \dfrac{n}{2n+1}$.
Q5 — Method of Differences & Telescoping Series · hard · numerical
The sum to infinity $\dfrac{1}{1 \cdot 3} + \dfrac{1}{3 \cdot 5} + \dfrac{1}{5 \cdot 7} + \cdots$ is:
A. $\dfrac14$
B. $1$
C. $\dfrac13$
D. $\dfrac12$ ✓ Correct
Solution: Partial sum $\dfrac{n}{2n+1} \to \dfrac12$ as $n \to \infty$.
Q6 — Method of Differences & Telescoping Series · hard · numerical
The sum to infinity $\sum_{r=1}^{\infty} \dfrac{1}{r(r+2)}$ is:
A. $\dfrac23$
B. $\dfrac12$
C. $1$
D. $\dfrac34$ ✓ Correct
Solution: $\dfrac{1}{r(r+2)} = \tfrac12\left(\dfrac1r - \dfrac{1}{r+2}\right)$; the sum telescopes to $\tfrac12\left(1 + \tfrac12\right) = \dfrac34$.
Q7 — Method of Differences & Telescoping Series · hard · numerical
The sum to infinity $\sum_{r=1}^{\infty} \dfrac{1}{r(r+1)(r+2)}$ is:
A. $\dfrac13$
B. $\dfrac18$
C. $\dfrac14$ ✓ Correct
D. $\dfrac12$
Solution: The $r$-th term $= \tfrac12\left(\dfrac{1}{r(r+1)} - \dfrac{1}{(r+1)(r+2)}\right)$; telescoping gives $\tfrac12 \cdot \tfrac12 = \dfrac14$.
Q8 — Method of Differences & Telescoping Series · medium · theory
In the method of differences, if the $r$-th term can be written as $T_r = V_r - V_{r-1}$, then $\sum_{r=1}^{n} T_r$ equals:
A. $V_n \cdot V_0$
B. $V_n + V_0$
C. $V_n - V_0$ ✓ Correct
D. $V_0 - V_n$
Solution: The intermediate terms cancel (telescope), leaving $V_n - V_0$.
Q9 — Method of Differences & Telescoping Series · hard · numerical
The value of $\sum_{r=1}^{4} r \cdot r!$ is:
A. $100$
B. $120$
C. $119$ ✓ Correct
D. $96$
Solution: Using $r \cdot r! = (r+1)! - r!$: the sum telescopes to $5! - 1 = 119$.
Q10 — Method of Differences & Telescoping Series · hard · numerical
The sum $\dfrac{1}{\sqrt1 + \sqrt2} + \dfrac{1}{\sqrt2 + \sqrt3} + \cdots + \dfrac{1}{\sqrt{99} + \sqrt{100}}$ is:
A. $9$ ✓ Correct
B. $10$
C. $\sqrt{99}$
D. $99$
Solution: Rationalising, each term $= \sqrt{r+1} - \sqrt{r}$; the sum telescopes to $\sqrt{100} - \sqrt1 = 9$.
Q11 — Method of Differences & Telescoping Series · medium · theory
Rationalising, $\dfrac{1}{\sqrt{r} + \sqrt{r+1}}$ equals:
A. $\dfrac{1}{\sqrt{r+1} - \sqrt{r}}$
B. $\sqrt{r} + \sqrt{r+1}$
C. $\sqrt{r+1} - \sqrt{r}$ ✓ Correct
D. $\sqrt{r} - \sqrt{r+1}$
Solution: Multiplying by the conjugate $\dfrac{\sqrt{r+1} - \sqrt{r}}{\sqrt{r+1} - \sqrt{r}}$ gives $\sqrt{r+1} - \sqrt{r}$.
Q12 — Method of Differences & Telescoping Series · medium · numerical
The sum $\sum_{r=1}^{n} \log\left(1 + \dfrac1r\right)$ equals:
A. $\log(n) - 1$
B. $\log(n+1)$ ✓ Correct
C. $\log n$
D. $n \log 2$
Solution: $\log\dfrac{r+1}{r}$ telescopes: $\sum = \log(n+1) - \log 1 = \log(n+1)$.
Q13 — Method of Differences & Telescoping Series · medium · numerical
The $n$-th term of the series $2, 5, 10, 17, 26, \dots$ (differences $3, 5, 7, \dots$ in A.P.) is:
A. $n^2 + 2$
B. $2n^2 - 1$
C. $n^2 + n$
D. $n^2 + 1$ ✓ Correct
Solution: Since the differences are in A.P., $T_n$ is quadratic; $T_n = n^2 + 1$ fits all terms.
Q14 — Method of Differences & Telescoping Series · hard · numerical
The $n$-th term of the series $3, 5, 9, 17, 33, \dots$ (differences $2, 4, 8, \dots$ in G.P.) is:
A. $3^{n-1}$
B. $2^{n-1} + 2$
C. $2^n - 1$
D. $2^n + 1$ ✓ Correct
Solution: Differences form a G.P.: $T_n = 3 + (2 + 4 + \cdots + 2^{n-1}) = 3 + (2^n - 2) = 2^n + 1$.
Q15 — Method of Differences & Telescoping Series · hard · numerical
The sum to infinity $\sum_{r=1}^{\infty} \dfrac{2r+1}{r^2 (r+1)^2}$ is:
A. $\dfrac34$
B. $\dfrac12$
C. $2$
D. $1$ ✓ Correct
Solution: The $r$-th term $= \dfrac{1}{r^2} - \dfrac{1}{(r+1)^2}$; telescoping to infinity gives $1$.
Q16 — Method of Differences & Telescoping Series · hard · numerical
The sum to infinity $\dfrac{1}{2 \cdot 5} + \dfrac{1}{5 \cdot 8} + \dfrac{1}{8 \cdot 11} + \cdots$ is:
A. $\dfrac16$ ✓ Correct
B. $\dfrac13$
C. $\dfrac12$
D. $\dfrac19$
Solution: Each term $= \tfrac13\left(\dfrac{1}{3r-1} - \dfrac{1}{3r+2}\right)$; the sum telescopes to $\tfrac13 \cdot \tfrac12 = \dfrac16$.
Q17 — Method of Differences & Telescoping Series · medium · numerical
The sum $\sum_{r=1}^{n} \dfrac{1}{(r+1)(r+2)}$ equals:
A. $\dfrac{n}{n+2}$
B. $1 - \dfrac{1}{n+2}$
C. $\dfrac12 - \dfrac{1}{n+2}$ ✓ Correct
D. $\dfrac{1}{n+2}$
Solution: $\dfrac{1}{(r+1)(r+2)} = \dfrac{1}{r+1} - \dfrac{1}{r+2}$; telescoping gives $\dfrac12 - \dfrac{1}{n+2}$.
Q18 — Method of Differences & Telescoping Series · hard · numerical
The sum $\sum_{r=1}^{n} \dfrac{1}{(3r-2)(3r+1)}$ equals:
A. $\dfrac{n}{3n+1}$ ✓ Correct
B. $\dfrac{3n}{3n+1}$
C. $\dfrac{n}{3n-2}$
D. $\dfrac{1}{3n+1}$
Solution: Each term $= \tfrac13\left(\dfrac{1}{3r-2} - \dfrac{1}{3r+1}\right)$; telescoping gives $\tfrac13\left(1 - \dfrac{1}{3n+1}\right) = \dfrac{n}{3n+1}$.
Q19 — Method of Differences & Telescoping Series · hard · numerical
The value of $\sum_{r=1}^{n} r(r+1)$ using $V_r = \dfrac{r(r+1)(r+2)}{3}$ (method of differences) for $n = 4$ is:
A. $30$
B. $50$
C. $40$ ✓ Correct
D. $20$
Solution: Using $V_r - V_{r-1} = r(r+1)$: $S_4 = \dfrac{4 \cdot 5 \cdot 6}{3} = 40$ (check: $2+6+12+20 = 40$).
Q20 — Method of Differences & Telescoping Series · medium · theory
The "$V_n$ method" for summing a series works best when the $r$-th term can be expressed as:
A. a difference $V_r - V_{r-1}$ of consecutive terms of some sequence ✓ Correct
B. a product of two G.P.s
C. a single power of $r$
D. a constant
Solution: When $T_r = V_r - V_{r-1}$, the sum telescopes, leaving only boundary terms.