(B) The sum of the squares of the first $n$ natural numbers is given by the formula: $\sum_{k=1}^{n} k^{2} = \frac{n(n+1)(2n+1)}{6}$.For $n = 10$,we s

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(B) The sum of the squares of the first $n$ natural numbers is given by the formula: $\sum_{k=1}^{n} k^{2} = \frac{n(n+1)(2n+1)}{6}$.For $n = 10$,we substitute the value into the formula:$1^{2}+2^{2}+3^{2}+\cdots+10^{2} = \frac{10(10+1)(2 \times 10+1)}{6}$.$= \frac{10 \times 11 \times 21}{6}$.$= \frac{2310}{6} = 385$.