Question

Find the sum to n-term
$3+15+35+63+\_\_\_\_\_\_\_$

Answer 1

Given sum

$3+9+15+35+63+...\dots .$

or

$3(1+3+5+7+9+....)$

the sum inside the brackets is sum of A.P

first term of this A.P is $a=1$

and common difference is $d=3-1=2$

sum of n terms of this A.P is given by

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

hence sum of n terms is

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

or

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

hence sum of n terms is $3n^2$