If 1 n + 1 C n n + 1 n C n - 1 n + . . . . + 1 2 C 1 n + C 0 n = 1023 10 , t h e n n i s e q u a l t o [2023]

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Solution

Q.
If $\frac{1}{n + 1} {}^{n}{C_{n}} + \frac{1}{n} {}^{n}{C_{n - 1}} + . . . . + \frac{1}{2} {}^{n}{C_{1}} + {}^{n}{C_{0}} = \frac{1023}{10}, t h e n n i s e q u a l t o$ [2023]

1 8

2 9

3 6

4 7

Ans.
(2)

$\frac{1}{n + 1} {}^{n}{C_{n}} + \frac{1}{n} {}^{n}{C_{n - 1}} + \dots + {}^{n}{C_{0}}$

$= \sum_{r = 0}^{n} \frac{{}^{n}{C_{r}}}{r + 1} = \frac{1}{n + 1} \sum_{r = 0}^{n} {}^{n + 1}{C_{r + 1}} = \frac{1}{n + 1} (2^{n + 1} - 1) = \frac{1023}{10} = \frac{2^{10} - 1}{10}$

$So, n + 1 = 10 \Rightarrow n = 9$

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