Question

The value of $1^2-2^2+3^2-4^2+.+99^2-100^2$ is :

• A. 5050
• B. 5075
• C. -5050
• D. -5075

Answer 1

Answer:

C. -5050

Sum of squares of n even numbers

$2^2+4^2+6^2+8^2+....++(2n)^2$

$Σ(2n)^2=\frac{2n(n+1)(2n+1)}{3}$

Sum of squares of n odd numbers

1²+3²+5² + .......+(2n-1)²

$Σ(2n-1)^2=\frac{n(2n+1)(2n-1)}{3}$

$1^2-2^2+3^2-4^2+....+99^2-100^2$

$=(1^2+3^2+..99^2)-(2^2+4^2+...+100^2)$

Number of odd = 50 and number of even = 50

sum of squares of odd=$\frac{n(2n+1)(2n-1)}{3}$

$=\frac{50(2\times 50+1)(2\times50-1)}{3}=\frac{50\times101\times99}{3}$

$=166650$

sum of squares of even = $\frac{2n(n+1)(2n+1)}{3}$

$=\frac{2 \times 50(50+1)(2\times 50+1)}{3}=\frac{100(51)(101)}{3}=100(17)(101)$

$=171700$

$(1^2+3^2+..99^2)-(2^2+4^2+...+100^2)=166650-171700=-5050$