The sum $1^2 + 2^2 + \cdots + n^2$ equals:
A$\dfrac{n(n+1)(2n+1)}{6}$
B$\dfrac{n(n+1)}{2}$
C$\left[\dfrac{n(n+1)}{2}\right]^2$
D$\dfrac{n(2n+1)}{6}$
Answer & Solution
Correct answer: A. $\dfrac{n(n+1)(2n+1)}{6}$
The standard special-series result: Σk² from 1 to n = n(n+1)(2n+1)/6.
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