Question

The value of $1^2+3^2+5^2+............+25^{2}is:$

• A. 2925
• B. 1469
• C. 1728
• D. 1456

Answer 1

Answer: (A)

Consider $1^2+3^2+5^2+.........+25^2$
$n^{th}term{T}_n={\left(2n-1\right)}^2,n=1,.......13$
$Now,S_n=\sum _{n=1}^{13}{T}_n=\sum _{n=1}^{13}{\left(2n-1\right)}^2$
$=\sum _{n=1}^{13}4n^2+\sum _{n=1}^{13}1-\sum _{n=1}^{13}4n$
$=4\sum n^2+13-4\sum n$
$=4\left[\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right]+13-4\frac{n\left(n+1\right)}{2}$
Put $n=13$, we get
$S_n=26\times 14\times 9+13-26\times 14$
= $3276+13-364$
= $2925$.