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JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Binomial Theorem

If

\frac{1}{n + 1}^{n} C_{n} + \frac{1}{n}^{n} C_{n - 1} + \dots + \frac{1}{2}^{n} C_{1} +^{n} C_{0} = \frac{1023}{10}

then n is equal to

(A)6
(B)9✓ Correct
(C)8
(D)7

Answer: (B)

Solution

Using the identity

\frac{1}{r + 1} (r n) = \frac{1}{n + 1} (r + 1 n + 1)

, we have

r = 0 \sum n \frac{( r n )}{r + 1} = \frac{1}{n + 1} r = 0 \sum n (r + 1 n + 1) = \frac{1}{n + 1} (2^{n + 1} - 1) .

Setting this equal to

\frac{1023}{10}

:

\frac{2 ^{n + 1} - 1}{n + 1} = \frac{1023}{10} .

Trying

n = 9

:

\frac{2 ^{10} - 1}{10} = \frac{1024 - 1}{10} = \frac{1023}{10}

. Thus

n = 9

.

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Single correct option

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