Question

Find the sum of the series $\sum_{r=0}^{n} (-1)^r . ^nC_r \left[ \frac{1}{2^r} + \frac{3}{2^{2r}} + \frac{7}{2^{3r}} + \frac{15}{2^{4r}} + ... up to n terms \right]$

• A. 0
• B. $-\frac{(2^n-1)^n}{2^{n^2}}$
• C. $\frac{(2^n-1)^n}{2^{n^2}}$
• D. 1

Answer 1

Answer: (B)

$-\frac{(2^n-1)^n}{2^{n^2}}$ for $n \ge 1$, and $0$ for $n=0$. Assuming $n$ is a positive integer for the context of such problems.

Answer 2

Let the given series be $S$. The expression is: $S = \sum_{r=0}^{n} (-1)^r . ^nC_r \left[ \frac{1}{2^r} + \frac{3}{2^{2r}} + \frac{7}{2^{3r}} + \frac{15}{2^{4r}} + ... \text{up to n terms} \right]$ The sum inside the bracket, $B_r$, is $\sum_{k=1}^{n} \frac{2^k-1}{2^{kr}}$. This can be written as a telescoping sum: $B_r = \sum_{k=1}^{n} \left(\frac{1}{2^{(k-1)r}} - \frac{1}{2^{kr}}\right) = 1 - \frac{1}{2^{nr}} = 1 - \left(\frac{1}{2^n}\right)^r$. Substituting back: $S = \sum_{r=0}^{n} (-1)^r . ^nC_r \left[ 1 - \left(\frac{1}{2^n}\right)^r \right]$ $S = \sum_{r=0}^{n} (-1)^r . ^nC_r - \sum_{r=0}^{n} (-1)^r . ^nC_r \left(\frac{1}{2^n}\right)^r$ The first sum is $(1-1)^n = 0$ for $n \ge 1$, and $1$ for $n=0$. The second sum is $\left(1 - \frac{1}{2^n}\right)^n$. For $n \ge 1$, $S = 0 - \left(1 - \frac{1}{2^n}\right)^n = -\left(\frac{2^n-1}{2^n}\right)^n = -\frac{(2^n-1)^n}{2^{n^2}}$. For $n=0$, $S = 1 - (1-1)^0 = 1 - 1 = 0$.