Question

Let n be a positive integer. Let
\(A=\sum_{k=0}^{n}(-1)^k\,{}^nC_k\left[\left(\frac12\right)^k+\left(\frac34\right)^k+\left(\frac78\right)^k+\left(\frac{15}{16}\right)^k+\left(\frac{31}{32}\right)^k\right]\).
If \(63A=1-\frac{1}{2^{30}}\), then n is equal to _____________.

Hint:

Use \(\sum(-1)^k\,{}^nC_k r^k=(1-r)^n\) for each ratio, then sum the resulting geometric series in \(2^{-n}\).

Answer 1

Correct Answer: 6

\(\sum_{k=0}^n(-1)^k\,{}^nC_kr^k=(1-r)^n\). Applying this to \(r=\frac12,\frac34,\frac78,\frac{15}{16},\frac{31}{32}\):
\(A=\left(\frac12\right)^n+\left(\frac14\right)^n+\left(\frac18\right)^n+\left(\frac1{16}\right)^n+\left(\frac1{32}\right)^n\).
Let \(x=2^{-n}\): \(A=x+x^2+x^3+x^4+x^5=\frac{x(1-x^5)}{1-x}\).
Try \(n=6\): \(x=2^{-6}\), \(A=\frac{2^{-6}(1-2^{-30})}{1-2^{-6}}=\frac{1-2^{-30}}{2^6-1}=\frac{1-2^{-30}}{63}\), so \(63A=1-\frac{1}{2^{30}}\). Since \(A\) is strictly decreasing in n, \(n=6\) is the only solution.
Answer: 6