Question

Find the sum to n terms of the series 3+15+35+63+.... .

Answer 1

Given series is 3+15+35+63+ ....

Here, the difference between the successive terms are 15-3=12, 35-15=20, 63-35=28, clearly these difference are in AP.

So, we use the difference method to find its nth term.

Let $T_n$ be the nth term and $S_n$ denotes the sum to n terms of the given series.

Then, $S_n=3+15+35+63+...+a_{n-1}+a_n$ ...(i)

Also, $S_n=3+15+35+...+a_{n-2}+a_{n-1}+a_n$ ...(ii)

On subtracting Eq. (ii) from Eq. (i), we get

$0=3+(12+20+28+...+a_n-a_{n-1}-a_n)$

$\Rightarrow a_n=3+\frac{n-1}{2}[2\times 12+(n-1-1\left)8\right]$

$=3+\frac{n-1}{2}(24+8n-16)$

$=3+(n-1)(4n+4)=4n^2-1$

$\therefore S_n=\sum a_n=\sum (4n^2-1)=4\sum n^2-\sum 1=\frac{4n(n+1)(2n+1)}{6}-n$

$=\frac{4n(n+1)(2n+1)-6n}{6}=\frac{n}{6}\left[4\right(n+1\left)\right(2n+1)-6]$

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

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