Question

The sum upto $(2n+1)$ terms of the series $a^2-(a+d{)}^2+(a+2d{)}^2-(a+3d{)}^2+....$ is

• A. $a^2+3n{d}^2$
• B. $a^2+2nad+n(n-1)d^2$
• C. $a^2+3nad+n(n-1)d^2$
• D. $a^2+2nad+n(2n+1)d^2$

Answer 1

The correct option is D $a^2+2nad+n(2n+1)d^2$

We can write the sum upto $(2n+1)$ terms as
$[a+(a+d\left)\right](-d)+\left[\right(a+2d)+(a+3d\left)\right](-d)+....\left[\right(a+(2n-2)d)+(a+(2n-1d](-d)+(a+2nd{)}^2$
$=(-d)[a+(a+d)+(a+2d)+.....+a+(2n-1\left)d\right]+(a+2nd{)}^2$
$=(-d)\frac{2n}{2}\{a+a+(2n-1\left)d\right\}+(a+2nd{)}^2$
$=-2nad-n(2n-1)d^2+a^2+4n\left(ad\right)+4n^2{d}^2$
$=a^2+2nad+n(2n+1)d^2$
Hence, option D is correct.