Question

Find the sum of the series whose nth term is :

(i) $2n^3+3n^{3-1}$

(ii) $n^3-3^n$

(iii) $n^{(}n+1\left)\right(n+4)$

(iv)$(2n-1{)}^2$

Answer 1

We have, $T_n=2n^3+3n^2-1$ Let $S_n$ denote the sum of n terms of the series whose nth term is $T_n$. Then, $S_n={\sum }_{k-1}^n{T}_k={\sum }_{k-1}^n(2k^3+3k^{2}1)={\sum }_{k-1}^{n}2k^3+{\sum }_{k-1}^{n}3k^2-{\sum }_{k-1}^{n}1=2{\sum }_{k-1}^n{k}^3+3{\sum }_{k-1}^n{k}^2-{\sum }_{k-1}^{n}1=2{\left[\frac{n(n+1)}{2}\right]}^2+3\left[\frac{n(n+1)(2n+1)}{6}\right]-n=2{\left[\frac{n(n+1)}{2}\right]}^2+3\left[\frac{n(n+1)(2n+1)}{6}\right]-n=\frac{2}{4}\left[n\right(n+1){]}^2+\frac{n(n+1)(2n+1)}{2}-n=\frac{\left[n\right(n+1){]}^2+n(n+1\left)\right(2n+1)}{2}-n=\frac{\left[n\right(n+1){]}^2+n(n+1\left)\right(2n+1)-2n}{2}=\frac{n}{2}[n^3+n+2n^2+2n^2+3n-1]=\frac{n}{2}[n(n+1{)}^2+(n+1\left)\right(2n+1)-2]Here,S_n=\frac{n}{2}(n^3+4n^2+4n)-1$

(ii) We have, $T_n=n^3-3^n$ Let $S_n$ denote the sum of n terms of the series whose nth term is $T_n$ $S_n={\sum }_{k-1}^n{T}_k={\sum }_{k-1}^n(k^3-3^k)={\sum }_{k-1}^n{k}^3-{\sum }_{k-1}^{n}3k{S}_n={\sum }_{k-1}^n{k}^3-{\sum }_{k-1}^{n}3k={\left[\frac{n(n+1)}{2}\right]}^2-(3^1+3^2...+3^n)=\frac{n^2(n+1{)}^2}{4}-3\left(\frac{3^n-1}{3-1}\right)=\frac{n^2(n+1{)}^2}{4}-\frac{3}{2}(3^n-1)Hence,S_n={\left[\frac{n(n+1)}{2}\right]}^2-\frac{3}{2}(3^n-1)$ (iii) We have, $T_n=n(n+1)(n+4)=(n^2+n)(n+4)=n^3+5n^2+4n$ Let $S_n$ denote the sum of n terms of the series not in L.C.M. whose nth term is $T_n$. Then, $S_n={\sum }_{k-1}^n{T}_k={\sum }_{k-1}^n(k^3+5k^2+4k)={\sum }_{k-1}^n{k}^3+{\sum }_{k-1}^{n}5k^2+{\sum }_{k-1}^{n}4k={\sum }_{k-1}^n{k}^3+5{\sum }_{k-1}^n{k}^2+4{\sum }_{k-1}^{n}k\Rightarrow S_n={\sum }_{k-1}^n{k}^3+5{\sum }_{k-1}^n{k}^2+4{\sum }_{k-1}^{n}k={\left[\frac{n(n+1)}{2}\right]}^2+5\left[\frac{n(n+1)(2n+1)}{6}\right]+\frac{4n(n+1)}{2}=\frac{1}{4}\left[n\right(n+1){]}^2+\frac{5n(n+1)(2n+1)}{6}+2n(n+1)=\frac{3\left[n\right(n+1){]}^2+10n(n+1\left)\right(2n+1)+24n(n+1)}{12}=\frac{n(n+1)}{12}[3n(n+1)+10(2n+1)+24]=\frac{n(n+1)}{12}[3n^2+3n+20n+10+24]=\frac{n(n+1)}{12}[3n^2+23n+34]Hence,S_n=\frac{n}{12}(n+1\left)\right(3n^2+23n+34)$ (iv) We have, $T_n$=(2n-$1{)}^2$ $=(2n{)}^2+1-2\times 2n\times 1=4n^2+1-4n=4n^2-4n+1\therefore T_n=4n^2-4n+1$ Let $S_n$ denote the sum of n terms of the series whose nth term is $T_n$. Then, $S_n={\sum }_{k-1}^n{T}_k={\sum }_{k-1}^{n}4k^2-{\sum }_{k-1}^{n}4k+{\sum }_{k-1}^{n}1\Rightarrow S_n=4{\sum }_{k-1}^n{k}^2-4{\sum }_{k-1}^{n}k+{\sum }_{k-1}^{n}1=4\left[\frac{n(n+1)(2n+1)}{6}\right]-4\frac{n(n+1)}{2}+n=\frac{2}{3}n(n+1)(2n+1)-2n(n+1)+n=\frac{2}{3}n(n+1)(2n+1)-2n^2-2n+n=\frac{2}{3}n(n+1)(2n+1)-2n^2-n=\frac{2}{3}(n+1)(2n+1)-n(2n+1)=\frac{2n(n+1)(2n+1)-3n(2n+1)}{3}=\frac{n}{3}(2n+1)\left[2\right(n+1)-3]=\frac{n}{3}(2n+1)(2n+2-3)=\frac{n}{3}(2n+1)(2n-1)Hence,S_n=\frac{n}{3}(2n+1)(2n-1)$