BINOMIAL SERIES, Vinod Singh ~ Kolkata
Find the sum of the series
\sum_{r=0}^{n} (-1)^r \binom{n}{r} \big( \frac{1}{2^r}+\frac{3^r}{2^{2r}}+\frac{7^r}{2^{3r}}+ \dots to m terms
\big)
Given series is equal to
\sum_{r=0}^{n} (-1)^r \binom{n}{r} \sum_{k=1}^{m} \frac{(2^k -1)^r}{2^{kr}}
= \sum_{r=0}^{n} \sum_{k=1}^{m} (-1)^r \binom{n}{r} \frac{(2^k -1)^r}{2^{kr}}
= \sum_{r=0}^{n} \sum_{k=1}^{m} (-1)^r \binom{n}{r} \lambda^r where
\lambda = \frac{(2^k -1)}{2^{k}}
= \sum_{k=1}^{m} \sum_{r=0}^{n} (-1)^r \binom{n}{r} \lambda^r Interchanging the summation
= \sum_{k=1}^{m} (1- \lambda)^n , Substituting the value of
\lambda we have,
= \sum_{k=1}^{m} \frac{1}{2^{nk}} = \frac{1}{2^n} \frac{\bigg(1- \big(\frac{1}{2^n}\big)^m \bigg)}{1 - \frac{1}{2^n} }
= \frac{1}{2^n} \frac{2^n(2^{nm}-1)}{2^{nm}(2^n-1)} = \frac{2^{nm}-1}{2^{nm}(2^n-1)}
