Question

Evaluate the following:
(i) $\sum _{n=1}^{11}(2+3^n)$

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

(iii) $\sum _{n=2}^{10}{4}^n$

Answer 1

(i)
$$\begin{gathered} S_{11}={\sum }_{n=1}^{11}\left(2+3^n\right) \\ \Rightarrow S_{11}={\sum }_{n=1}^{11}2+{\sum }_{n=1}^{11}{3}^n \\ \Rightarrow S_{11}=2\times 11+\left(3+3^2+3^3+...+3^{11}\right) \\ =22+3\left(\frac{3^{11}-1}{3-1}\right) \\ =22+\left(\frac{177147-1}{2}\right) \\ =22+265719 \\ =265741 \end{gathered}$$

(ii)
$$\begin{gathered} S_n=\sum _{k=1}^n\left(2^k+3^{k-1}\right) \\ =\sum _{k=1}^n{2}^k+\sum _{k=1}^n{3}^{k-1} \\ =\left(2+4+8+...+2^n\right)+\left(1+3+9+...+3^n\right) \\ =2\left(\frac{2^n-1}{2-1}\right)+1\left(\frac{3^n-1}{3-1}\right) \\ =\frac{1}{2}\left(2^{n+2}-4+3^n-1\right) \\ =\frac{1}{2}\left(2^{n+2}+3^n-5\right) \end{gathered}$$

(iii)
$$\begin{gathered} \sum _{n=2}^{10}{4}^n=4^2+4^3+4^4+...+4^{10} \\ =16+64+256+...+4^{10} \\ =16\left(\frac{4^9-1}{4-1}\right)=\frac{16}{3}\left(4^9-1\right) \end{gathered}$$