sum of n terms

Question

find sum of n terms

1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + ...................

 

can anyone explain me this  how to solve this

Answer 1

JUST TAKE ODD NUMBER ON ONE SIDE AND EVEN NUMBER ON OTHER SIDE LIKE THIS

(1^2 + 3^2 + ..(2N-1)^2) - (2^2 + 4^2 + ...(2N)^2)

NOW,

SUM OF SQUARE OF ODD NATURAL NUMBER = N(2N+1)(2N-1)/3

SUM OF SQUARE OF EVEN NATURAL NUMBER = 2N(N+1)(2N+1)/3

TOTAL = ODD SERIES - EVEN SERIES

            = - N (2N+1)

Comments

when n is even   sum  =  -n(n+1)/2

when n ii s odd sum =  n(n+1)/2

 

its a question from rd sharma

---

(1+2)(1-2) + (3+4)(3-4)+(5-6)(5+6) +.........................((2n-1)-2n) ((2n -1) + 2n)

-[3+7+11+.........(4n-1)]

-(n/2)*(4n+2) = -n*(2n+1)