Let \(n\) be a positive integer. Let \(A=\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\left[\left(\frac{1}{2}\right)^{k}+\left(\frac{3}{4}\right)^{k}+\left(\frac{7}{8}\right)^{k}+\left(\frac{15}{16}\right)^{k}+\left(\frac{31}{32}\right)^{k}\right]\). If \(63 A=1-\frac{1}{2^{30}}\), then \(n\) is equal to ______

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Binomial Theorem

Let

n

be a positive integer. Let

A = \sum_{k = 0}^{n} (- 1)^{k} (k n) [(\frac{1}{2})^{k} + (\frac{3}{4})^{k} + (\frac{7}{8})^{k} + (\frac{15}{16})^{k} + (\frac{31}{32})^{k}]

. If

63 A = 1 - \frac{1}{2 ^{30}}

, then

n

is equal to ______

Answer: 6

By the binomial theorem

\sum_{k = 0}^{n} (- 1)^{k} (k n) x^{k} = (1 - x)^{n}

, applied to each of the five brackets,

A = (\frac{1}{2})^{n} + (\frac{1}{4})^{n} + (\frac{1}{8})^{n} + (\frac{1}{16})^{n} + (\frac{1}{32})^{n}

This is a geometric sum of five terms with first term

2^{- n}

and ratio

2^{- n}

, so

A = \frac{2 ^{- n} ( 1 - 2 ^{- 5 n} )}{1 - 2 ^{- n}} = \frac{1 - 2 ^{- 5 n}}{2 ^{n} - 1}

n = 6

the denominator is

2^{6} - 1 = 63

and the numerator is

1 - 2^{- 30}

, so

63 A = 1 - \frac{1}{2 ^{30}}

exactly as required. Every one of the five powers

2^{- n}, 2^{- 2 n}, \dots, 2^{- 5 n}

decreases strictly as

n

grows, so

A

is strictly decreasing and no other positive integer satisfies the condition, giving

n = 6

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Integer answer