Question

Find the sum $\sum _{k=1}^n(2^k+3^{k-1})$

Answer 1

$\Rightarrow {\sum }_{k=1}^n{2}^k=2^1+2^2+2^3+......2^n$
$=\frac{2(2^n-1)}{2-1}=2^{n+1}-2$
${\sum }_{k=1}^n{3}^{k-1}=3^0+3^1+3^2+......3^{n-1}$
$=\frac{3^0(3^n-1)}{3-1}=\frac{3^n}{2}-\frac{1}{2}$
${\sum }_{k=1}^n(2^k+3^{k-1})=2^{n+1}-2+\frac{3^n}{2}-\frac{1}{2}$
$=2^{n+1}+\frac{3^n}{2}-\frac{5}{2}$