$n (n + 1)^{3} < 8 (1^{3} + 2^{3} + 3^{3} + \dots \dots + n^{3})$

True

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False

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Solution

True
Apply A.M. of $1^{3}, 2^{3}, \dots \dots n^{3}$ is > cube of A.M. of 1, 2, 3, ...... n and
$1 + 2 + 3 + \dots \dots + n = \frac{n}{2} (n + 1)$

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Similar questions

n {(n + 1)}^{3} < 8 (1^{3} + 2^{3} + 3^{3} + . . . . + n^{3})

1^{3} + 2^{3} + 3^{3} + . . . . . . + n^{3} = {(\frac{n (n + 1)}{2})}^{2}

1^{3} + 2^{3} + 3^{3} + . . . . . . + 21^{3}

1^{3} + 2^{3} + 3^{3} + . . . . . . . + n^{3} = {(\frac{n (n + 1)}{2})}^{2}

$1^{3} + 2^{3} + 3^{3} + \dots + n^{3} = {[\frac{n (n + 1)}{2}]}^{2}$

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

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