Observe that
$1^{3} = 1$ , $2^{3} = 3 + 5$ , $3^{3} = 7 + 9 + 11$ , $4^{3} = 13 + 15 + 17 + 19$ .
Then $n^{3}$ as a similar series is

[2 {\frac{n (n - 1)}{2} + 1} - 1] + [2 {\frac{(n + 1) n}{2} + 1} + 1] + . . . . + [2 {\frac{n (n - 1)}{2} + 1} + 2 n - 3]

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(n^{2} + n + 1) + (n^{2} + n + 3) + (n^{2} + n + 5) + . . . + (n^{2} + 3 n - 1)

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(n^{2} - n + 1) + (n^{2} - n + 3) + (n^{2} - n + 5) + . . . + (n^{2} + n - 1)

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none of these

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Solution

The correct option is C $(n^{2} - n + 1) + (n^{2} - n + 3) + (n^{2} - n + 5) + . . . + (n^{2} + n - 1)$
Given
$2^{3} = 3 + 5, 3^{3} = 7 + 9 + 11, 4^{3} = 13 + 15 + 17 + 19$
$\Rightarrow 2^{3} = (2^{2} - 1) + (2^{2} + 1), 3^{3} = (3^{2} - 2) + (3^{2}) + (3^{2} + 2), 4^{3} = (4^{2} - 3) + (4^{2} - 1) + (4^{2} + 1) + (4^{2} + 3)$
or $2^{3} = (2^{2} - 2 + 1) + (2^{2} - 2 + 3), 3^{3} = (3^{2} - 3 + 1) + (3^{2} - 3 + 3) + (3^{2} - 3 + 5), 4^{3} = (4^{2} - 4 + 1) + (4^{2} - 4 + 3) + (4^{2} - 4 + 5) + (4^{2} - 4 + 7)$
Thus, $n^{3} = (n^{2} - n + 1) + (n^{2} - n + 3) + (n^{2} - n + 5) + . . . + (n^{2} - n + (2 n - 1))$
$∴ C$ is correct.

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Similar questions

Q. Given

\infty \sum n = 1 \frac{1}{n^{2}} = \frac{π^{2}}{6}

: the value of

\infty \sum n = 1 \frac{1 + 3 + 5 + . . . . . . + (2 n - 1)}{1^{3} + 2^{3} + 3^{3} + . . . . . . . . n^{3}}

is:

Q. Show that

\frac{1}{1 -} \frac{4}{5 -} \frac{9}{13 -} \frac{64}{25 -} \dots \frac{{(n^{2} - 1)}^{2}}{n^{2} + {(n + 1)}^{2}} = \frac{(n + 1) (n + 2) (2 n + 3)}{6}

$= {lim}_{n \to \infty} [\frac{1}{n} + \frac{1}{\sqrt{n^{2} + n}} + \frac{1}{\sqrt{n^{2} + 2 n}} + \dots + \frac{1}{\sqrt{n^{2} + (n - 1) n}}]$ is equal to [RPET 2000]

Q. The sum of series

{sec}^{- 1} \sqrt{2} + {sec}^{- 1} \frac{\sqrt{10}}{3} + {sec}^{- 1} \frac{\sqrt{50}}{7} + . . . + {sec}^{- 1} \sqrt{\frac{(n^{2} + 1) (n^{2} - 2 n + 2)}{(n^{2} - n + 1)^{2}}}

{lim}_{n \to \infty} \frac{1. n^{2} + 2 (n - 1)^{2} + 3 (n - 2)^{2} + . . . + n {.1}^{2}}{1^{3} + 2^{3} + 3^{3} + . . . + n^{3}}

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Observe that13=1, 23=3+5, 33=7+9+11, 43=13+15+17+19.Then n3 as a similar series is

Observe that
$1^{3} = 1$ , $2^{3} = 3 + 5$ , $3^{3} = 7 + 9 + 11$ , $4^{3} = 13 + 15 + 17 + 19$ .
Then $n^{3}$ as a similar series is