Evaluate the following
n∑i=1(3/2(i−1))n\sum i=1 (3/2^(i-1))n∑i=1(3/2(i−1))
The aim is to compute the sum: n∑i=0nin\sum_{i=0}^nin∑i=0ni. The sum ∑i=0ni\sum_{i=0}^ni∑i=0ni is an arithmetic progression. It is a well-known formula. n∑i=0ni=n(0+1+2+…+n)=n(n(n+1)2)=n2(n+1)2.n\sum_{i=0}^ni=n(0+1+2+\ldots+n)=n\left(\frac{n(n+1)}{2}\right)=\frac{n^2(n+1)}{2}.n∑i=0ni=n(0+1+2+…+n)=n(2n(n+1))=2n2(n+1).
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