JEE Main 2013 (Online) 25th AprilQuestion 1
The value of $1^{2}+3^{2}+5^{2}+\ldots+25^{2}$ is :
Sequence and Series · Standard Sums (Σ formulas)
Solution
Formulas used. (1) $\sum_{k=1}^{n}(2k-1)^2=\frac{n(2n-1)(2n+1)}{3}$Step 1 - Number of terms. $1,3,\dots,25$: $n=13$. Step 2 - Formula. $$\frac{13\times25\times27}{3}=2925 .$$ Check. All squares up to $26^2$ minus even squares: $\dfrac{26\cdot27\cdot53}{6}-4\cdot\dfrac{13\cdot14\cdot27}{6}=6201-3276=2925$. Answer: (A)