# Thread: sums .

1. ## sums .

Deduce the sum of the series .

(1) $\displaystyle 1^2-2^2+3^2-4^2+...+(2n-1)^2-(2n)^2+(2n+1)^2$

(2) $\displaystyle 25^2-26^2+27^2-28^2+...+49^2-50^2$

THanks .

2. Originally Posted by thereddevils
Deduce the sum of the series .

(1) $\displaystyle 1^2-2^2+3^2-4^2+...+(2n-1)^2-(2n)^2+(2n+1)^2$

$\displaystyle a^2 -b^2 =(a+b)(a-b)$

Hence
$\displaystyle 1^2-2^2+3^2-4^2+...+(2n-1)^2-(2n)^2+(2n+1)^2$

$\displaystyle = (1+2)(1-2) +~(2+3)(2-3) + ~......... ( ~( 2n-1) + 2n)~) ( ~( 2n-1)- 2n)~) + (2n+1)^2$

$\displaystyle = -1[ 3 +7 +...........(4n-1) ] + (2n+1)^2$

$\displaystyle =-1[\frac{n}{2}(2\times 3 + (n-1)4)] + (2n+1)^2$............................Using AP

$\displaystyle = -1[ n(2n+1) ] (2n+1)^2$

$\displaystyle = -1[2n^2 +n] + 4n^2 + 1 +4n$

$\displaystyle = 2n^2 + 3n +1$

$\displaystyle = (2n+1)(n+1)$

(2)
Now for this try it the same way

$\displaystyle = -1[ 51 +55 +...........(4\times 25 -1) ]$

$\displaystyle =-1[\frac{25}{2}(2\times 51 + (13-1)4)]$

Go ahead and ask incase of difficulty