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Binomial Theorem Part 3

Section: Mathematics

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Question : 14 of 53

Marks: +1, -0

Let

C_{r}

denote the coefficient of

x^{r}

in the binomial expansion of

(1+x)^{n}, n \in N, 0 \leq r \leq n

P_n=C_0-C_1+\;\frac{2^2}{3}C_2-\;\frac{2^3}{4}C_3+\ldots+\;\frac{(-2)^n}{n+1}C_n

, then the value of

\sum\limits_{n=1}^{25}\;\frac{1}{P_{2n}}

[22 Jan 2026 Shift 2]

675
580
525
650

Solution:

Concept:

Use the identity

\frac{{}^nC_r}{r+1} = \frac{1}{n+1}{}^{n+1}C_{r+1}

to simplify the sum

P_n

.

Explanation:

P_n = \sum_{r=0}^n \frac{{}^nC_r(-2)^r}{r+1}

= \frac{1}{n+1}\sum_{r=0}^n {}^{n+1}C_{r+1}(-2)^r

= \frac{-1}{2(n+1)}\sum_{r=0}^n {}^{n+1}C_{r+1}(-2)^{r+1}

= \frac{-1}{2(n+1)}\left[(1-2)^{n+1} - 1\right]

using binomial expansion.

Thus

P_n = \frac{1}{2(n+1)}[1 - (-1)^{n+1}]

.

For even

n = 2n

:

P_{2n} = \frac{1}{2(2n+1)}[1 - (-1)^{2n+1}] = \frac{1}{2(2n+1)}[1 - (-1)] = \frac{1}{2n+1}

.

So

\frac{1}{P_{2n}} = 2n+1

.

Sum from

n=1

to

25

:

\sum_{n=1}^{25} (2n+1) = 3+5+\dots+51

.

Number of terms = 25, average =

\frac{3+51}{2}=27

, sum =

25 \times 27 = 675

.

Answer:

675

(Option A)

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