Question

Find the sum of $n$ terms of the series
$\frac{2a^2-1}{a},4a-\frac{3}{a},\frac{6a^2-5}{a},.....$

Answer 1

series $⟹\frac{2a^2-1}{a},4a-\frac{3}{a},\frac{6a^2-5}{a},........nterms$
$=2a-\frac{1}{a},2\left(2a\right)-\frac{1+2}{a},3\left(2a\right)-\frac{1+2\left(2\right)}{a},........nterms$
$T_1=2a-\frac{1}{a}$
$T_2=2\left(2a\right)-\frac{(1+2)}{a}$
$T_3=3\left(2a\right)-\frac{1+2\left(2\right)}{a}$
$T_4=4\left(2a\right)-\frac{1+2\left(3\right)}{a}$
$\therefore T_n=n\left(2a\right)-\frac{1+2(n-1)}{a}$
$T_n=2an-\frac{(2n-1)}{a}$
$\sum T_n=2a\sum n-\frac{1}{a}\sum (2n-1)$
$=2a\frac{n(n+1)}{2}-\frac{1}{a}\times 2\frac{n(n+1)}{2}+\frac{1}{a}\left(n\right)$
$=an(n+1)-\frac{n(n+1)}{a}+\left[\frac{n}{a}\right]$$=\frac{a^{2}n(n+1)-n(n+1)+n}{a}$
$=\frac{a^{2}n(n+1)-n^2}{a}=\frac{n^2[a^2-1]+a^{2}n}{a}$