Question

Find the sum to $n$ terms of the series $1^2-2^2+3^2-4^2+5^2-6^2+...$

Answer 1

$1^2-2^2+3^2-4^2+5^2-6^2+....n$ terms

Case 1 : Let n be an even integer 2m

Then

$1^2-2^2+3^2-4^2+5^2-6^2+....$upto 2m terms

$=(1^2-2^2)+(3^2-4^2)+(5^2-6^2)+....$ upto m terms

$=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+....m$ terms

$=-3-7-11-....$upto m terms

$=-(3+7+11+...uptomterms)$

$=\frac{m}{2}[2\times 3+(m-1)4]$

$=-m(2m+1)$ .........Equation no. (1)

$=-\frac{n}{2}(n+1)(\because n=2m)$

Thus, if n is an even integer then the sum of the given series upto n terms$=-\frac{n}{2}(n+1)$

Case 2: let n be an odd integer = 2m+1

Then,

$=1^2-2^2+3^2-4^2+5^2-6^2+....$ to (2m

1) terms

$=(1^2-2^2)+(3^2-4^2)+(5^2-6^2)+....$ to m terms$+{(2m+1)}^2$

$=-m(2m+1)+{(2m+1)}^2\left[UsingEquation\right(1\left)\right]$

$=(2m+1)+(m+1)$

$=n(\frac{n-1}{2}+1)(\because n=2m+1andm=\frac{n-1}{2})$

$=\frac{n(n+1)}{2}$

Hence if n is an odd integer then the sum of the series upto n terms $=\frac{n(n+1)}{2}$