Question

Find the sum of $n$ terms of the series $5+11+19+29+41+..$

Answer 1

$s=1+5+11+19+29+41+\cdots +an$ $-----\left(1\right)$

$s=1+5+11+19+29+\cdots +an-1+an$ $----\left(2\right)$

Subtracting $\left(2\right)$ from $\left(1\right)$ we get

$0=1+\left(4+6+8+10+12+\cdots +\cdots an-1\right)-an$

$\Rightarrow an=1+\left(4+6+8+10+12+\cdots +an-1\right)$

$\Rightarrow an=1+2+\left(n-2\right)2$

$\Rightarrow an=1+2+2n-4$

$\Rightarrow an=2n-1$

Therefore sum of $n$ terms of the given series

$=\sum _1^{n}an$

$=\sum _1^n\left(2n-1\right)$

$=2\sum _1^{n}n-\sum _1^{n}1$

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

$=n^2+n-n$

$=n^2$