Question

Statement-1: The sum of first $n$ terms of the series $1^2-2^2+3^2-4^2+5^2\dots \dots \dots \text{ can be }=\pm \frac{n(n+1)}{2}.$
Statement-2: Sum of first $n$ natural numbers is $\frac{n(n+1)}{2}$.

• A. Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
• B. Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
• C. Statement-1 is true, statement-2 is false.
• D. Statement-1 is false, statement-2 is true. .

Answer 1

Answer: (A)

$(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+\dots \dots ..\text{ if }$ n is odd
$-1-2-3-4-5-6\dots \dots (n-2)-(n-1)+n^2$
$=-\frac{(n-1)n}{2}+n^2=\frac{2n^2-n^2+n}{2}=\frac{n^2+n}{2}=\frac{n(n+1)}{2}<br/>$
when $n$ is even
$=-1-2-3-4\dots \dots ..(n-1)-n=-\frac{n(n+1)}{2}$