Let n be a positive integer. Let A = ∑ _ k = 0 ^ n (- 1) ^ k × Ck n [ ((1)/(2)) ^ k + ((3)/(4)) ^ k + ((7)/(8)) ^ k + ((15)/(16)) ^ k + ((31)/(32)) ^ k ] . If 63A = 1 - 1 2 ^ 30 , then n is equal to ______ .

Correct answer: 0 A = ∑ _ k = 0 ^ n Ck n [ (- (1)/(2)) ^ k + ((- 3)/(4)) ^ k + ((- 7)/(8)) ^ k + ((- 15)/(16)) ^ k + ((- 31)/(32)) ^ k ] A = (1 - (1)/(2)) ^ n + (1 - (3)/(4)) ^ n + (1 - (7)/(8)) ^ n + (1 - (15)/(16)) ^ n + (1 - (31)/(32)) ^ n A = 1 2 ^ n + 1 4 ^ n + 1 8 ^ n + 1 16 ^ n + 1 32 ^ n A = 1 2 ^ n ( 1 - ( 1 2 ^ n ) ^ 5 1 - 1 2 ^ n ) ⇒ A = (1 - 1 2 ^ 5n ) ( 2 ^ n - 1) ( 2 ^ n - 1)A = 1 - 1 2 ^ 5n , Given, 63A = 1 - 1 2 ^ 30 Clearly 5n = 30 n = 6

JEE Main 202116 MarchEvening Shift

Exam: JEE Main
Year: 2021
Date: 16 March
Shift: Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2021 · 16 March · Evening Shift · Mathematics · Binomial Theorem

Let n be a positive integer. Let A = ∑ _ k = 0 ^ n (- 1) ^ k × Ck n...

Let

n

be a positive integer. Let

A = \mathop{∑}_{k = 0}^{n}{\left(- 1\right)}^{k} \times Ck n \left[{\left(\dfrac{1}{2}\right)}^{k} + {\left(\dfrac{3}{4}\right)}^{k} + {\left(\dfrac{7}{8}\right)}^{k} + {\left(\dfrac{15}{16}\right)}^{k} + {\left(\dfrac{31}{32}\right)}^{k}\right]

. If

63A = 1 - \dfrac{1}{{2}^{30}},

then

n

is equal to ______ .

Correct answer

Step-by-step solution

A = \mathop{∑}_{k = 0}^{n} Ck n \left[{\left(- \dfrac{1}{2}\right)}^{k} + {\left(\dfrac{- 3}{4}\right)}^{k} + {\left(\dfrac{- 7}{8}\right)}^{k} + {\left(\dfrac{- 15}{16}\right)}^{k} + {\left(\dfrac{- 31}{32}\right)}^{k}\right] $$ A = {\left(1 - \dfrac{1}{2}\right)}^{n} + {\left(1 - \dfrac{3}{4}\right)}^{n} + {\left(1 - \dfrac{7}{8}\right)}^{n} + {\left(1 - \dfrac{15}{16}\right)}^{n} + {\left(1 - \dfrac{31}{32}\right)}^{n} $$ A = \dfrac{1}{{2}^{n}} + \dfrac{1}{{4}^{n}} + \dfrac{1}{{8}^{n}} + \dfrac{1}{{16}^{n}} + \dfrac{1}{{32}^{n}} $$ A = \dfrac{1}{{2}^{n}}\left(\dfrac{1 - {\left(\dfrac{1}{{2}^{n}}\right)}^{5}}{1 - \dfrac{1}{{2}^{n}}}\right) ⇒ A = \dfrac{\left(1 - \dfrac{1}{{2}^{5n}}\right)}{\left({2}^{n} - 1\right)} $\left({2}^{n} - 1\right)A = 1 - \dfrac{1}{{2}^{5n}},$

Given, $63A = 1 - \dfrac{1}{{2}^{30}}$

Clearly $5n = 30$

$n = 6$

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