Limn - 1-13-n3-4-23-n3- - 1-2n is equal to

Let S=lim_π→∞ 1/1^3+n^3+4/2^3+n^3+⋯ +1/2n =lim_π→∞ 1/1^3+n^3+4/2^3+n^3+⋯ + n^2/n^3+n^3 ∴ S=lim_π→∞∑_r=1^n r^2/r^3+n^3=lim_π→∞r^2/n^3 ( r^3/n^3+1 ) = lim_π→∞∑_r ...

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Byju's Answer

Standard XIII

Mathematics

Definite Integral as Limit of Sum

limn →∞ 1/13+...

Question

${lim}_{n \to \infty} \frac{1}{1^{3} + n^{3}} + \frac{4}{2^{3} + n^{3}} + \dots + \frac{1}{2 n}$ is equal to

\frac{1}{3} l o g_{e} 3

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\frac{1}{3} l o g_{e} 2

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\frac{1}{3} l o g_{e} \frac{1}{3}

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$N o n e o f t h e s e$

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Solution

The correct option is B $\frac{1}{3} l o g_{e} 2$
$L e t S = {lim}_{π \to \infty} \frac{1}{1^{3} + n^{3}} + \frac{4}{2^{3} + n^{3}} + \dots + \frac{1}{2 n}$
$= {lim}_{π \to \infty} \frac{1}{1^{3} + n^{3}} + \frac{4}{2^{3} + n^{3}} + \dots + \frac{n^{2}}{n^{3} + n^{3}}$
$∴ S = {lim}_{π \to \infty} \sum_{r = 1}^{n} \frac{r^{2}}{r^{3} + n^{3}} = {lim}_{π \to \infty} \frac{r^{2}}{n^{3} (\frac{r^{3}}{n^{3}} + 1)}$
$= {lim}_{π \to \infty} \sum_{r = 1}^{n} \frac{1}{n} . \frac{{(\frac{r}{n})}^{2}}{[1 + {(\frac{r}{n})}^{3}]}$
Applying the formula, we get $A = \int_{0}^{1} \frac{x^{2}}{1 + x^{3}} d x$
$= \frac{1}{3} \int_{0}^{1} \frac{3 x^{2}}{1 + x^{3}} d x = \frac{1}{3} [l o g_{e} (1 + x^{3})]_{0}^{1} = \frac{1}{3} l o g_{e} 2$

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Similar questions

{lim}_{n \to \infty} \frac{1}{1^{3} + n^{3}} + \frac{4}{2^{3} + n^{3}} + \dots + \frac{1}{2 n}

is equal to

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

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

is equal to :

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limn→∞ 113+n3+423+n3+⋯+12n is equal to

The correct option is B 13 loge 2Let S=limπ→∞ 113+n3+423+n3+⋯+12n =limπ→∞ 113+n3+423+n3+⋯+n2n3+n3 ∴ S=limπ→∞∑nr=1r2r3+n3=limπ→∞r2n3(r3n3+1) =limπ→∞∑nr=1 1n.(rn)2[1+(rn)3] Applying the formula, we get A=∫10 x21+x3dx =13∫10 3x21+x3dx=13[loge(1+x3)]10=13loge2

${lim}_{n \to \infty} \frac{1}{1^{3} + n^{3}} + \frac{4}{2^{3} + n^{3}} + \dots + \frac{1}{2 n}$ is equal to