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) $S_{11}={\sum }_{n=1}^{11}(2+3^n)\Rightarrow S_{11}=(2+3^1)+(2+3^2)+(2+3^3)+...+(2+3^{11})\Rightarrow S_{11}=2\times 11+3^1+3^2+3^3+....+3^{11}=22+\frac{3(3^{11}-1)}{(3-1)}=22+\frac{3(3^{11}-1)}{2}=\frac{44+3(177147-1)}{2}=\frac{44+3\left(177146\right)}{2}=265741$

So,

${\sum }_{n=1}^{11}(2+3^n)=265741\left(ii\right)S_n={\sum }_{k=1}^n(2^k+3^{k-1})=(2+3^0)+(2^2+3)+(2^3+3^2)+....+(2^n+3^{n-1})=(2+2^2+2^3+...+2^n)+(3^0+3^1+3^2+...+3^{n-1})=(s_n+s_m)S_n\Rightarrow a=2,n=n,r=\frac{2^2}{2}=2S_n=\frac{a(r^n-1)}{r-1}=\frac{2(2^n-1)}{(2-1)}=2(2^n-1)Also,S_m=S_{n-1}a=1,r=3,n=n-1S_{n-1}=\frac{1(3^{n-1}-1)}{3-1}=\frac{1}{2}(3^n-1)\therefore {\sum }_{k=1}^n(2^k+3^{k-1})=2(2^n-1)+\frac{1}{2}(3^n-1)=\frac{1}{2}[2^{n+2}+3^n-4-1]=\frac{1}{2}[2^{n+2}+3^n-5]$

(iii) ${\sum }_{n=2}^{10}{4}^n=4^2+4^3+4^4+...+4^{10}a=4^2,r=\frac{4^3}{4}=4,n=9S_{10}=\frac{a(r^9-1)}{1-r}=\frac{4^2(4^9-1)}{4-1}=\frac{1}{3}[4^{11}-16]=\frac{16}{3}[4^9-1]$