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The sum $1^2 + 2^2 + 3^2 + \cdots + 10^2$ equals:
A$3025$, the square of the simple sum $1 + 2 + \cdots + 10$
B$100$, the squared count of terms in the sum here
C$55$, the simple sum of the first ten positive integers
D$385$, since $\sum k^2 = n(n+1)(2n+1)/6 = 10\cdot 11\cdot 21/6$
Answer & Solution
Correct answer: D. $385$, since $\sum k^2 = n(n+1)(2n+1)/6 = 10\cdot 11\cdot 21/6$
$\sum_{k=1}^{10} k^2 = 10\cdot 11\cdot 21/6 = 385$.
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