Evaluate - k - 1 n 2 k - 3 k - 1

∑_k=1^n(2^k+3^k 1) =∑_k=1^n2^k+∑_k=1^n3^k 1 =2+4+8+.............+1+3+9+27 =2[2^n 11]+1[3^n 12] =2^n 1 2+3^n 12 =2^n 1+3^n2 52 ...

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Summation by Sigma Method

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Question

Evaluate $\sum_{k = 1}^{n} (2^{k} + 3^{k - 1})$

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Solution

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