(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:$1^{2}+2^{2}+3^{2}+\cdots+10^{2} = \frac{10(10+1)(2 \times 10+1)}{6}$.$= \frac{10 \times 11 \times 21}{6}$.$= \frac{2310}{6} = 385$.

Competitive Quantitative Aptitude Progression and Sequence Progression and Sequence

Medium

$(1^{2}+2^{2}+3^{2}+\cdots+10^{2})$ is equal to:

A
$380$
B
$385$
C
$390$
D
$392$

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Progression and Sequence

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Progression and Sequence

If $x, y, z \in R^+$ such that $x + y + z = 4$,then the maximum possible value of $xyz^2$ is -

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$\left[\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\cdots+\frac{1}{99 \times 100}\right]$ is equal to

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The $6^{th}$ term of a $G.P.$ is $32$ and its $8^{th}$ term is $128$,then the common ratio of the $G.P.$ is

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$(1^{2}+2^{2}+3^{2}+\cdots+10^{2})$ is equal to:

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