If (1)/(n + 1) ^ n C _ n + (1)/(n) ^ n C _ n - 1 + . . . + (1)/(2) ^ n C _ 1 + ^ n C _ 0 = (1023)/(10) then n is equal to

Correct answer: A. 9 Given that (1)/(n + 1) ^ n C _ n + (1)/(n) ^ n C _ n - 1 + . . . + (1)/(2) ^ n C _ 1 + ^ n C _ 0 = (1023)/(10) ⇒ ∑ _ r = 0 ^ n (1)/(r + 1) ^ n C _ r = (1023)/(10)(∴ ^ n + 1 C _ r + 1 = (n + 1)/(r + 1) ^ n C _ r ) ⇒ ∑ _ r = 0 ^ n (1)/(n + 1) ^ n + 1 C _ r + 1 = (1023)/(10) ⇒ (1)/(n + 1)[ ^ n + 1 C _ 1 + ^ n + 1 C _ 2 + . . . + ^ n + 1 C _ n + 1 ] = (1023)/(10) We know that C0 n + C1 n + C2 n + . . . . . . . + Cn n = 2 ^ n ⇒ 2 ^ n + 1 - 1 n + 1 = (1023)/(10) = 2 ^ 10 - 1 10 ⇒ n + 1 = 10 ⇒ n = 9 Hence this is the correct option.

JEE Main 202312 AprilMorning Shift

Exam: JEE Main
Year: 2023
Date: 12 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2023 · 12 April · Morning Shift · Mathematics · Binomial Theorem

If (1)/(n + 1) ^ n C _ n + (1)/(n) ^ n C _ n - 1 + . . . + (1)/(2)...

If $\dfrac{1}{n + 1}{}^{n}{C}_{n} + \dfrac{1}{n}{}^{n}{C}_{n - 1} + . . . + \dfrac{1}{2}{}^{n}{C}_{1}{+}^{n}{C}_{0} = \dfrac{1023}{10}$ then $n$ is equal to

Options

A $9$
B $8$
C $7$
D $6$

Correct answer

A. $9$

Step-by-step solution

Given that $\dfrac{1}{n + 1}{}^{n}{C}_{n} + \dfrac{1}{n}{}^{n}{C}_{n - 1} + . . . + \dfrac{1}{2}{}^{n}{C}_{1}{+}^{n}{C}_{0} = \dfrac{1023}{10}$

$⇒ \mathop{∑}_{r = 0}^{n}\dfrac{1}{r + 1}{}^{n}{C}_{r} = \dfrac{1023}{10}\left(∴ {}^{n + 1}{C}_{r + 1} = \dfrac{n + 1}{r + 1}{}^{n}{C}_{r}\right)$

$⇒ \mathop{∑}_{r = 0}^{n}\dfrac{1}{n + 1}{}^{n + 1}{C}_{r + 1} = \dfrac{1023}{10}$

$⇒ \dfrac{1}{n + 1}\left[{}^{n + 1}{C}_{1}{+}^{n + 1}{C}_{2} + . . . {+}^{n + 1}{C}_{n + 1}\right] = \dfrac{1023}{10}$

We know that $C0 n + C1 n + C2 n + . . . . . . . + Cn n = {2}^{n}$

$⇒ \dfrac{{2}^{n + 1} - 1}{n + 1} = \dfrac{1023}{10} = \dfrac{{2}^{10} - 1}{10}$

$⇒ n + 1 = 10$

$⇒ n = 9$

Hence this is the correct option.

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