JEE PYQ: Binomial Theorem Question 3
Question 3 - 2021 (16 Mar Shift 2)
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 $63A = 1 - \frac{1}{2^{30}}$, then $n$ is equal to ____.
Show Answer
Answer: 6
Solution
$A = \frac{1}{2^n} + \frac{1}{4^n} + \frac{1}{8^n} + \frac{1}{16^n} + \frac{1}{32^n}$. Geometric series gives $63A = 1 - 1/2^{5n}$, so $2^n - 1 = 63$, $n = 6$.
BETA
