Let n be a positive integer. Let A = _ k =0 ^ n (-1)^ k n _ C _ k… - Vidyadip

MCQ

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

A
$12$
B
$8$
✓
$6$
D
$16$

✓

Answer

Correct option: C.

$6$

c
$A=\sum_{k=0}^{n}{ }^{n} C_{k}\left[\left(-\frac{1}{2}\right)^{k}+\left(\frac{-3}{4}\right)^{k}+\left(\frac{-7}{8}\right)^{k}+\left(\frac{-15}{16}\right)^{k}+\left(\frac{-37}{32}\right)^{k}\right]$

$A=\left(1-\frac{1}{2}\right)^{n}+\left(1-\frac{3}{4}\right)^{n}+\left(1-\frac{7}{8}\right)^{n}+\left(1-\frac{15}{16}\right)^{n}+\left(1-\frac{31}{32}\right)^{n}$

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

$A=\frac{1}{2^{n}}\left(\frac{1-\left(\frac{1}{2^{n}}\right)^{5}}{1-\frac{1}{2^{n}}}\right) \Rightarrow A=\frac{\left(1-\frac{1}{2^{5 n}}\right)}{\left(2^{n}-1\right)}$

$\left(2^{n}-1\right) A=1-\frac{1}{2^{5 n}},$ Given $63 A =1-\frac{1}{2^{30}}$

Clearly $5 n=30$

$n=6$

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