Question

$a+\frac{1}{3},3a-\frac{1}{6},5a+\frac{1}{12}+....$ to $2p$ terms.

Answer 1

Sum of first $n$ odd natural numbers is given by
$n^2$
Given series $S=a+\frac{1}{3},3a-\frac{1}{6},5a+\frac{1}{12}+....$ to $2p$ terms, can be divided into two parts
Let $S_1=a+3a+5a+............2p$ terms
Clearly, this represents sum of first $2p$ odd natural numbers
Then, $S_1=(2p{)}^{2}a=4p^{2}a$
$S_2=\frac{1}{3}-\frac{1}{6}+\frac{1}{12}-\frac{1}{24}..........2p$ terms
Common ratio $=\frac{-1}{2}$
$S_2=\frac{\frac{1}{3}(1-{\left[\frac{-1}{2}\right]}^{2p})}{1+\frac{1}{2}}$
$S_2=\frac{2(1-\frac{1}{4^p})}{9}$
$S=S_1+S_2=4p^{2}a+\frac{2(1-\frac{1}{4^p})}{9}$