Question

Find the sum of the series whose nth term is:
(i) 2n² − 3n + 5
(ii) 2n³ + 3n² − 1
(iii) n³ − 3ⁿ
(iv) n (n + 1) (n + 4)
(v) (2n − 1)²

Answer 1

Let $T_n$ be the nth term of the given series.

Thus, we have:

(i)
$T_n=2n^2-3n+5$

Let $S_n$ be the sum of n terms of the given series.

Now,
$S_n=\sum _{k=1}^n{T}_k$

$$\begin{gathered} \Rightarrow S_n=\sum _{k=1}^n\left(2k^2-3k+5\right) \\ \Rightarrow S_n=\underset{k=1}{\overset{n}{2\sum }}{k}^2-3\sum _{k=1}^{n}k+\sum _{k=1}^{n}5 \\ \Rightarrow S_n=\frac{2n\left(n+1\right)\left(2n+1\right)}{6}-\frac{3n\left(n+1\right)}{2}+5n \\ \Rightarrow S_n=\frac{2n\left(n+1\right)\left(2n+1\right)-9n\left(n+1\right)+30n}{6} \\ \Rightarrow S_n=\frac{\left(2n^2+2n\right)\left(2n+1\right)-9n^2-9n+30n}{6} \\ \Rightarrow S_n=\frac{4n^3+4n^2+2n^2+2n-9n^2-9n+30n}{6} \\ \Rightarrow S_n=\frac{4n^3-3n^2+23n}{6} \\ \Rightarrow S_n=\frac{n\left(4n^2-3n+23\right)}{6} \end{gathered}$$

(ii)
$T_n=2n^3+3n^2-1$

Let $S_n$ be the sum of n terms of the given series.

Now,
$S_n=\sum _{k=1}^n{T}_k$

$$\begin{gathered} \Rightarrow S_n=\sum _{k=1}^n\left(2k^3+3k^2-1\right) \\ \Rightarrow S_n=\underset{k=1}{\overset{n}{2\sum }}{k}^3+3\sum _{k=1}^n{k}^2-\sum _{k=1}^{n}1 \\ \Rightarrow S_n=\frac{2n^2{\left(n+1\right)}^2}{4}+\frac{3n\left(n+1\right)\left(2n+1\right)}{6}-n \\ \Rightarrow S_n=\frac{n^2{\left(n+1\right)}^2}{2}+\frac{n\left(n+1\right)\left(2n+1\right)}{2}-n \\ \Rightarrow S_n=\frac{n^2{\left(n+1\right)}^2+n\left(n+1\right)\left(2n+1\right)-2n}{2} \\ \Rightarrow S_n=\frac{n^2\left(n^2+1+2n\right)+\left(n^2+n\right)\left(2n+1\right)-2n}{2} \\ \Rightarrow S_n=\frac{\left(n^4+n^2+2n^3\right)+\left(2n^3+n^2+2n^2+n\right)-2n}{2} \\ \Rightarrow S_n=\frac{n^4+4n^2+4n^3-n}{2} \\ \Rightarrow S_n=\frac{n\left(n^3+4n+4n^2-1\right)}{2} \end{gathered}$$

(iii)
$T_n=n^3-3^n$

Let $S_n$ be the sum of n terms of the given series.

Now,
$S_n=\sum _{k=1}^n{T}_k$

$$\begin{gathered} \Rightarrow S_n=\sum _{k=1}^n\left(k^3-3^k\right) \\ \Rightarrow S_n=\sum _{k=1}^n{k}^3-\sum _{k=1}^n{3}^k \\ \Rightarrow S_n=\frac{n^2{\left(n+1\right)}^2}{4}-\left(3+3^2+3^3+3^4+...+3^n\right) \\ \Rightarrow S_n=\frac{n^2{\left(n+1\right)}^2}{4}-\left[\frac{3\left(3^n-1\right)}{3-1}\right] \\ \Rightarrow S_n=\frac{n^2{\left(n+1\right)}^2}{4}-\frac{3}{2}\left(3^n-1\right) \end{gathered}$$

(iv)
$T_n=n\left(n+1\right)\left(n+4\right)=\left(n^2+n\right)\left(n+4\right)=n^3+5n^2+4n$

Let $S_n$ be the sum of n terms of the given series.

Now,
$S_n=\sum _{k=1}^n{T}_k$

$$\begin{gathered} \Rightarrow S_n=\sum _{k=1}^n\left(k^3+5k^2+4k\right) \\ \Rightarrow S_n=\sum _{k=1}^n{k}^3+\underset{k=1}{\overset{n}{5\sum }}{k}^2+4\sum _{k=1}^{n}k \\ \Rightarrow S_n=\frac{n^2{\left(n+1\right)}^2}{4}+\frac{5n\left(n+1\right)\left(2n+1\right)}{6}+\frac{4n\left(n+1\right)}{2} \\ \Rightarrow S_n=\frac{n\left(n+1\right)}{2}\left[\frac{n\left(n+1\right)}{2}+\frac{5\left(2n+1\right)}{3}+4\right] \\ \Rightarrow S_n=\frac{n\left(n+1\right)}{12}\left[3n\left(n+1\right)+10\left(2n+1\right)+24\right] \\ \Rightarrow S_n=\frac{n\left(n+1\right)}{12}\left(3n^2+23n+34\right) \end{gathered}$$


(v)
$T_n={\left(2n-1\right)}^2$

Let $S_n$ be the sum of n terms of the given series.

Now,
$S_n=\sum _{k=1}^n{T}_k$

$$\begin{gathered} \Rightarrow S_n=\sum _{k=1}^n{\left(2k-1\right)}^2 \\ \Rightarrow S_n=\sum _{k=1}^n\left(4k^2+1-4k\right) \\ \Rightarrow S_n=\underset{k=1}{\overset{n}{4\sum }}{k}^2+\underset{k=1}{\overset{n}{\sum 1}}-4\sum _{k=1}^{n}k \\ \Rightarrow S_n=\frac{4n\left(n+1\right)\left(2n+1\right)}{6}+n-\frac{4n\left(n+1\right)}{2} \\ \Rightarrow S_n=\frac{n\left(n+1\right)}{2}\left[\frac{4\left(2n+1\right)}{3}-4\right]+n \\ \Rightarrow S_n=\frac{n\left(n+1\right)}{2}\left(\frac{8n+4-12}{3}\right)+n \\ \Rightarrow S_n=\frac{n\left(n+1\right)}{2}\left(\frac{8n-8}{3}\right)+n \\ \Rightarrow S_n=4n\left(n+1\right)\left(\frac{n-1}{3}\right)+n \\ \Rightarrow S_n=\frac{n\left(4n+4\right)\left(n-1\right)+3n}{3} \\ \Rightarrow S_n=\frac{n}{3}\left(4n^2+4n-4n-4+3\right) \\ \Rightarrow S_n=\frac{n}{3}\left(4n^2-1\right) \\ \Rightarrow S_n=\frac{n}{3}\left(2n-1\right)\left(2n+1\right) \end{gathered}$$