\lim_{n\rightarrow \infty }(\frac{(n+1)^{\frac{1}{3}}}{(n)^{\frac{4}{3}}})+\frac{(n+2)^{\frac{1}{3}}}{(n)^{\frac{4}{3}}}+............+\frac{(2n)^{\frac{1}{3}}}{(n)^{\frac{4}{3}}}) 

is equal to :

  • Option 1)

    \frac{3}{4}(2)^{\frac{4}{3}}-\frac{3}{4}

  • Option 2)

    \frac{4}{3}(2)^{\frac{4}{3}}

  • Option 3)

    \frac{3}{4}(2)^{\frac{4}{3}}-\frac{4}{3}

  • Option 4)

    \frac{4}{3}(2)^{\frac{3}{4}}

 

Answers (1)

\lim_{n\rightarrow \infty }(\frac{(n+1)^{\frac{1}{3}}}{(n)^{\frac{4}{3}}}+\frac{(n+2)^{\frac{1}{3}}}{(n)^{\frac{4}{3}}}+............+\frac{(2n)^{\frac{1}{3}}}{(n)^{\frac{4}{3}}})

=> \lim_{n\rightarrow \infty }\frac{1}{n}[(1+\frac{1}{n})^{\frac{1}{3}}+(1+\frac{2}{n})^{\frac{1}{3}}+..........+(1+\frac{n}{n})^{\frac{1}{3}}]

=> \frac{1}{n}\sum_{n=1}^{n}(1+\frac{r}{n})^{\frac{1}{3}}

=> \int_{0}^{1}(1+x)^{\frac{1}{3}}dx=\frac{3}{4}(2^{\frac{4}{3}}-1)

option (1) is correct.


Option 1)

\frac{3}{4}(2)^{\frac{4}{3}}-\frac{3}{4}

Option 2)

\frac{4}{3}(2)^{\frac{4}{3}}

Option 3)

\frac{3}{4}(2)^{\frac{4}{3}}-\frac{4}{3}

Option 4)

\frac{4}{3}(2)^{\frac{3}{4}}

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