Sequences And Series — Easy JEE math MCQ
The sum \(1^2 + 2^2 + 3^2 + \ldots + 15^2\) equals:
- A. 1120
- B. 1240
- C. 1360
- D. 1480
Solution
The sum of squares of first \(n\) natural numbers is:
\[
\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}
\]
For \(n = 15\):
\[
\sum_{k=1}^{15} k^2 = \frac{15 \times 16 \times 31}{6} = \frac{7440}{6} = 1240
\]
So the correct option is B.