Find the sum of the series whose nth term is:

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

Solution:

Let $T_{n}$ be the $n$th term of the given series.

Thus, we have:

(i) $T_{n}=2 n^{2}-3 n+5$

Let $S_{n}$ be the sum of $n$ terms of the given series.

Now,

$S_{n}=\sum_{k=1}^{n} T_{k}$

$\Rightarrow S_{n}=\sum_{k=1}^{n}\left(2 k^{2}-3 k+5\right)$

$\Rightarrow S_{n}=2 \sum_{k=1}^{n} k^{2}-3 \sum_{k=1}^{n} k+\sum_{k=1}^{n} 5$

$\Rightarrow S_{n}=\frac{2 n(n+1)(2 n+1)}{6}-\frac{3 n(n+1)}{2}+5 n$

$\Rightarrow S_{n}=\frac{2 n(n+1)(2 n+1)-9 n(n+1)+30 n}{6}$

$\Rightarrow S_{n}=\frac{\left(2 n^{2}+2 n\right)(2 n+1)-9 n^{2}-9 n+30 n}{6}$

$\Rightarrow S_{n}=\frac{4 n^{3}+4 n^{2}+2 n^{2}+2 n-9 n^{2}-9 n+30 n}{6}$

$\Rightarrow S_{n}=\frac{4 n^{3}-3 n^{2}+23 n}{6}$

$\Rightarrow S_{n}=\frac{n\left(4 n^{2}-3 n+23\right)}{6}$

(ii) $T_{n}=2 n^{3}+3 n^{2}-1$

Let $S_{n}$ be the sum of $n$ terms of the given series.

Now,

$S_{n}=\sum_{k=1}^{n} T_{k}$

$\Rightarrow S_{n}=\sum_{k=1}^{n}\left(2 k^{3}+3 k^{2}-1\right)$

$\Rightarrow S_{n}=2 \sum_{k=1}^{n} k^{3}+3 \sum_{k=1}^{n} k^{2}-\sum_{k=1}^{n} 1$

$\Rightarrow S_{n}=\frac{2 n^{2}(n+1)^{2}}{4}+\frac{3 n(n+1)(2 n+1)}{6}-n$

$\Rightarrow S_{n}=\frac{n^{2}(n+1)^{2}}{2}+\frac{n(n+1)(2 n+1)}{2}-n$

$\Rightarrow S_{n}=\frac{n^{2}(n+1)^{2}+n(n+1)(2 n+1)-2 n}{2}$

$\Rightarrow S_{n}=\frac{n^{2}\left(n^{2}+1+2 n\right)+\left(n^{2}+n\right)(2 n+1)-2 n}{2}$

$\Rightarrow S_{n}=\frac{\left(n^{4}+n^{2}+2 n^{3}\right)+\left(2 n^{3}+n^{2}+2 n^{2}+n\right)-2 n}{2}$

$\Rightarrow S_{n}=\frac{n^{4}+4 n^{2}+4 n^{3}-n}{2}$

$\Rightarrow S_{n}=\frac{n\left(n^{3}+4 n+4 n^{2}-1\right)}{2}$

(iii) $T_{n}=n^{3}-3^{n}$

Let $S_{n}$ be the sum of $n$ terms of the given series.

Now,

$S_{n}=\sum_{k=1}^{n} T_{k}$

$\Rightarrow S_{n}=\sum_{k=1}^{n}\left(k^{3}-3^{k}\right)$

$\Rightarrow S_{n}=\sum_{k=1}^{n} k^{3}-\sum_{k=1}^{n} 3^{k}$

$\Rightarrow S_{n}=\frac{n^{2}(n+1)^{2}}{4}-\left(3+3^{2}+3^{3}+3^{4}+\ldots+3^{n}\right)$

$\Rightarrow S_{n}=\frac{n^{2}(n+1)^{2}}{4}-\left[\frac{3\left(3^{n}-1\right)}{3-1}\right]$

$\Rightarrow S_{n}=\frac{n^{2}(n+1)^{2}}{4}-\frac{3}{2}\left(3^{n}-1\right)$

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

Let $S_{n}$ be the sum of $n$ terms of the given series.

Now,

$S_{n}=\sum_{k=1}^{n} T_{k}$

$\Rightarrow S_{n}=\sum_{k=1}^{n}\left(k^{3}+5 k^{2}+4 k\right)$

$\Rightarrow S_{n}=\sum_{k=1}^{n} k^{3}+5 \sum_{k=1}^{n} k^{2}+4 \sum_{k=1}^{n} k$

$\Rightarrow S_{n}=\frac{n^{2}(n+1)^{2}}{4}+\frac{5 n(n+1)(2 n+1)}{6}+\frac{4 n(n+1)}{2}$

$\Rightarrow S_{n}=\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{5(2 n+1)}{3}+4\right]$

$\Rightarrow S_{n}=\frac{n(n+1)}{12}[3 n(n+1)+10(2 n+1)+24]$

$\Rightarrow S_{n}=\frac{n(n+1)}{12}\left(3 n^{2}+23 n+34\right)$

(v) $T_{n}=(2 n-1)^{2}$

Let $S_{n}$ be the sum of $n$ terms of the given series.

Now,

$S_{n}=\sum_{k=1}^{n} T_{k}$

$\Rightarrow S_{n}=\sum_{k=1}^{n}(2 k-1)^{2}$

$\Rightarrow S_{n}=\sum_{k=1}^{n}\left(4 k^{2}+1-4 k\right)$

$\Rightarrow S_{n}=4 \sum_{k=1}^{n} k^{2}+\sum_{k=1}^{n} 1-4 \sum_{k=1}^{n} k$

$\Rightarrow S_{n}=\frac{4 n(n+1)(2 n+1)}{6}+n-\frac{4 n(n+1)}{2}$

$\Rightarrow S_{n}=\frac{n(n+1)}{2}\left[\frac{4(2 n+1)}{3}-4\right]+n$

$\Rightarrow S_{n}=\frac{n(n+1)}{2}\left(\frac{8 n+4-12}{3}\right)+n$

$\Rightarrow S_{n}=\frac{n(n+1)}{2}\left(\frac{8 n-8}{3}\right)+n$

$\Rightarrow S_{n}=4 n(n+1)\left(\frac{n-1}{3}\right)+n$

$\Rightarrow S_{n}=\frac{n(4 n+4)(n-1)+3 n}{3}$

$\Rightarrow S_{n}=\frac{n}{3}\left(4 n^{2}+4 n-4 n-4+3\right)$

$\Rightarrow S_{n}=\frac{n}{3}\left(4 n^{2}-1\right)$

$\Rightarrow S_{n}=\frac{n}{3}(2 n-1)(2 n+1)$

 

 

 

 

 

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now