Sum of the ` $n$ ' brackets of $(1) + (1 + 3) + (1 + 3 + 5) + . . .$ is

\frac{n (n + 1)}{2}

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\frac{n (n + 1) (2 n + 1)}{6}

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n^{2} (n + 1)^{2}

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n^{2} + n

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Solution

The correct option is B $\frac{n (n + 1) (2 n + 1)}{6}$
$S = Σ (Σ 2 r + 1)$
$S = Σ n^{2}$
$S = \frac{n (n + 1) (2 n + 1)}{6}$

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The sum of (n-1) terms of 1+(1+3)+(1+3+5)+.............is

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