Why does sum from n=1 to N of (V_n - V_(n-1)) telescope to V_N - V_0?

Writing out the terms: (V_1 - V_0) + (V_2 - V_1) + … + (V_N - V_(N-1)), every intermediate term V_1, V_2, …, V_(N-1) cancels out, leaving only V_N - V_0.

JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Sum of a telescoping series using partial fractions

Q: Why is the numerator -2 split as (2n - 5) - (2n - 3)?

Because (2n - 5) - (2n - 3) = -5 - (-3) = -2, which matches the numerator directly without introducing extra minus signs.

If a_n=(-2)/(4 n^2-16 n+15), then a_1+a_2+…+a_25 is equal to:

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What feels right?

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A52/147B49/138C50/141D51/144

Step-by-step solutionView

Correct answer

Factor the quadratic denominator into $(2 n - 5) (2 n - 3)$ , decompose $a_{n}$ into partial fractions as $\frac{1}{2 n - 3} - \frac{1}{2 n - 5}$ , and evaluate the telescoping sum to get $\frac{50}{141}$ .

Option analysis

Why each option works or fails

A · 52/147

Summing 26 terms instead of 25 terms gives $\frac{1}{3} + \frac{1}{2 (26) - 3} = \frac{1}{3} + \frac{1}{49} = \frac{52}{147}$ . Check the upper summation limit: for $n = 25$ , the remaining positive term is $\frac{1}{2 (25) - 3} = \frac{1}{47}$ , not $\frac{1}{49}$ .

B · 49/138

Summing only 23 terms or confusing the index offset results in an upper boundary of $\frac{1}{43}$ , giving $\frac{1}{3} - (- \frac{1}{43}) = \frac{46}{129}$ or similar incorrect offsets like $\frac{49}{138}$ . Carefully track index cancellation: $a_{n} = \frac{1}{2 n - 3} - \frac{1}{2 n - 5}$ . The surviving terms after 25 additions are precisely $- \frac{1}{2 (1) - 5} = \frac{1}{3}$ and $\frac{1}{2 (25) - 3} = \frac{1}{47}$ .

C · 50/141

The partial fraction decomposition $\frac{1}{2 n - 3} - \frac{1}{2 n - 5}$ telescopes, leaving $\frac{1}{3} + \frac{1}{47} = \frac{50}{141}$ . This is the correct answer.

D · 51/144

Summing 24 terms instead of 25 terms leaves $\frac{1}{3} + \frac{1}{2 (24) - 3} = \frac{1}{3} + \frac{1}{45} = \frac{16}{45} = \frac{51.2}{144}$ or an off-by-one error at the boundary resulting in $\frac{51}{144}$ . Ensure the evaluation includes all terms from $n = 1$ through $n = 25$ inclusive.

Reviewed route

Solution

StepWorking

01Given

General term is $a_{n} = \frac{- 2}{4 n^{2} - 16 n + 15}$ and we need to find $S = \sum_{n = 1}^{25} a_{n}$ .

02Goal

Evaluate $\sum_{n = 1}^{25} a_{n}$ in exact fractional form.

03Approach

Factor the quadratic denominator $4 n^{2} - 16 n + 15 = (2 n - 3) (2 n - 5)$ . Then decompose $a_{n}$ into partial fractions to set up a telescoping sum.

04Execute

Factor the denominator: $4 n^{2} - 16 n + 15 = (2 n - 3) (2 n - 5)$ . Notice the difference between the factors: $(2 n - 3) - (2 n - 5) = 2$ . Therefore, $- 2 = (2 n - 5) - (2 n - 3)$ . Rewrite $a_{n}$ : $a_{n} = \frac{(2 n - 5) - (2 n - 3)}{(2 n - 3) (2 n - 5)} = \frac{1}{2 n - 3} - \frac{1}{2 n - 5}$ .

05Execute

Let $V_{n} = \frac{1}{2 n - 3}$ . Then $V_{n - 1} = \frac{1}{2 (n - 1) - 3} = \frac{1}{2 n - 5}$ . So $a_{n} = V_{n} - V_{n - 1}$ , forming a direct telescoping sum: $\sum_{n = 1}^{25} a_{n} = \sum_{n = 1}^{25} (V_{n} - V_{n - 1}) = V_{25} - V_{0}$ . Evaluate the terms: $V_{25} = \frac{1}{2 (25) - 3} = \frac{1}{47}$ . $V_{0} = \frac{1}{2 (0) - 3} = - \frac{1}{3}$ . Hence, the sum is: $S = \frac{1}{47} - (- \frac{1}{3}) = \frac{1}{47} + \frac{1}{3} = \frac{3 + 47}{47 \times 3} = \frac{50}{141}$ .

✓Verify

Check for $N = 1$ : $a_{1} = \frac{- 2}{4 - 16 + 15} = \frac{- 2}{3}$ . Formula gives $V_{1} - V_{0} = - 1 - (- \frac{1}{3}) = - \frac{2}{3}$ . Correct. For $N = 25$ : $\frac{50}{141} \approx 0.3546$ . Option C matches precisely.

Hints that build this answer step by step

How does the denominator $4 n^{2} - 16 n + 15$ factor?
$(2 n - 5) (2 n - 3)$
What is the partial fraction decomposition of $a_{n} = \frac{- 2}{(2 n - 5) (2 n - 3)}$ ?
$\frac{1}{2 n - 3} - \frac{1}{2 n - 5}$
When telescoping $\sum_{n = 1}^{25} (\frac{1}{2 n - 3} - \frac{1}{2 n - 5})$ , which terms remain uncanceled?
$- \frac{1}{2 (1) - 5} + \frac{1}{2 (25) - 3}$
What is the evaluated value of $\frac{1}{3} + \frac{1}{47}$ ?
$\frac{50}{141}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Why is the numerator

- 2

split as

(2 n - 5) - (2 n - 3)

Because $(2 n - 5) - (2 n - 3) = - 5 - (- 3) = - 2$ , which matches the numerator directly without introducing extra minus signs.

Why does

\sum_{n = 1}^{N} (V_{n} - V_{n - 1})

telescope to

V_{N} - V_{0}

Writing out the terms: $(V_{1} - V_{0}) + (V_{2} - V_{1}) + \dots + (V_{N} - V_{N - 1})$ , every intermediate term $V_{1}, V_{2}, \dots, V_{N - 1}$ cancels out, leaving only $V_{N} - V_{0}$ .

Answer

Factor the quadratic denominator into $(2 n - 5) (2 n - 3)$ , decompose $a_{n}$ into partial fractions as $\frac{1}{2 n - 3} - \frac{1}{2 n - 5}$ , and evaluate the telescoping sum to get $\frac{50}{141}$ .

Why each option works or fails

A: 52/147 - Summing 26 terms instead of 25 terms gives $\frac{1}{3} + \frac{1}{2 (26) - 3} = \frac{1}{3} + \frac{1}{49} = \frac{52}{147}$ . Check the upper summation limit: for $n = 25$ , the remaining positive term is $\frac{1}{2 (25) - 3} = \frac{1}{47}$ , not $\frac{1}{49}$ .
B: 49/138 - Summing only 23 terms or confusing the index offset results in an upper boundary of $\frac{1}{43}$ , giving $\frac{1}{3} - (- \frac{1}{43}) = \frac{46}{129}$ or similar incorrect offsets like $\frac{49}{138}$ . Carefully track index cancellation: $a_{n} = \frac{1}{2 n - 3} - \frac{1}{2 n - 5}$ . The surviving terms after 25 additions are precisely $- \frac{1}{2 (1) - 5} = \frac{1}{3}$ and $\frac{1}{2 (25) - 3} = \frac{1}{47}$ .
C · correct: 50/141 - The partial fraction decomposition $\frac{1}{2 n - 3} - \frac{1}{2 n - 5}$ telescopes, leaving $\frac{1}{3} + \frac{1}{47} = \frac{50}{141}$ . This is the correct answer.
D: 51/144 - Summing 24 terms instead of 25 terms leaves $\frac{1}{3} + \frac{1}{2 (24) - 3} = \frac{1}{3} + \frac{1}{45} = \frac{16}{45} = \frac{51.2}{144}$ or an off-by-one error at the boundary resulting in $\frac{51}{144}$ . Ensure the evaluation includes all terms from $n = 1$ through $n = 25$ inclusive.

Step-by-step solution

Given: General term is $a_{n} = \frac{- 2}{4 n^{2} - 16 n + 15}$ and we need to find $S = \sum_{n = 1}^{25} a_{n}$ .
Goal: Evaluate $\sum_{n = 1}^{25} a_{n}$ in exact fractional form.
Approach: Factor the quadratic denominator $4 n^{2} - 16 n + 15 = (2 n - 3) (2 n - 5)$ . Then decompose $a_{n}$ into partial fractions to set up a telescoping sum.
Execute: Factor the denominator: $4 n^{2} - 16 n + 15 = (2 n - 3) (2 n - 5)$ . Notice the difference between the factors: $(2 n - 3) - (2 n - 5) = 2$ . Therefore, $- 2 = (2 n - 5) - (2 n - 3)$ . Rewrite $a_{n}$ : $a_{n} = \frac{(2 n - 5) - (2 n - 3)}{(2 n - 3) (2 n - 5)} = \frac{1}{2 n - 3} - \frac{1}{2 n - 5}$ .
Execute: Let $V_{n} = \frac{1}{2 n - 3}$ . Then $V_{n - 1} = \frac{1}{2 (n - 1) - 3} = \frac{1}{2 n - 5}$ . So $a_{n} = V_{n} - V_{n - 1}$ , forming a direct telescoping sum: $\sum_{n = 1}^{25} a_{n} = \sum_{n = 1}^{25} (V_{n} - V_{n - 1}) = V_{25} - V_{0}$ . Evaluate the terms: $V_{25} = \frac{1}{2 (25) - 3} = \frac{1}{47}$ . $V_{0} = \frac{1}{2 (0) - 3} = - \frac{1}{3}$ . Hence, the sum is: $S = \frac{1}{47} - (- \frac{1}{3}) = \frac{1}{47} + \frac{1}{3} = \frac{3 + 47}{47 \times 3} = \frac{50}{141}$ .
Verify: Check for $N = 1$ : $a_{1} = \frac{- 2}{4 - 16 + 15} = \frac{- 2}{3}$ . Formula gives $V_{1} - V_{0} = - 1 - (- \frac{1}{3}) = - \frac{2}{3}$ . Correct. For $N = 25$ : $\frac{50}{141} \approx 0.3546$ . Option C matches precisely.

Sum of a telescoping series using partial fractions

What feels right?

Factor the quadratic denominator into $(2 n - 5) (2 n - 3)$ , decompose $a_{n}$ into partial fractions as $\frac{1}{2 n - 3} - \frac{1}{2 n - 5}$ , and evaluate the telescoping sum to get $\frac{50}{141}$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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