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If \mathrm{[x]} denotes the greatest integer less than or equal to \mathrm{x}, then, \mathrm{\lim _{n \rightarrow \infty} \frac{\left[1^2 x\right]+\left[2^2 x\right]+\left[3^2 x\right]+\ldots .+\left[n^2 x\right]}{n^3}} equals
 

Option: 1

\mathrm{x / 2}
 


Option: 2

\mathrm{x / 3}
 


Option: 3

\mathrm{x / 6}
 


Option: 4

0


Answers (1)

best_answer

For any integer k, we have

\mathrm{ k^2 x-1<\left[k^2 x\right] \leq k^2 x+1 }

\mathrm{ \Rightarrow \quad \sum_{k=1}^n\left(k^2 x-1\right)<\sum_{k=1}^n\left[k^2 x\right] \leq \sum_{k=1}^n\left(k^2 x+1\right) }

\mathrm{ \Rightarrow \frac{1}{n^3} \sum_{k=1}^n\left(k^2 x-1\right)<\frac{1}{n^3} \sum_{k=1}^n\left[k^2 x\right] \leq \frac{1}{n^3} \sum_{k=1}^n\left(k^2 n+1\right) }

\mathrm{ \Rightarrow \frac{1}{n^3}\left(\frac{n(n+1)(2 n+1)}{6}\right) x-\frac{n}{n^3}<\frac{1}{n^3} \sum_{k=1}^n\left[k^2 x\right] }

\mathrm{ \leq\left(\frac{n(n+1)(2 n+1)}{6}\right) x+\frac{n}{n^3} }

\mathrm{ \Rightarrow\left(\frac{2 n^2+3 n+1}{6 n^2}\right) x-\frac{1}{n^2}<\frac{1}{n^3} \sum_{k=1}^n\left[k^2 x\right] \leq\left(\frac{2 n^2+3 n+1}{6 n^2}\right) x+\frac{1}{n^2} }


Taking limit as \mathrm{ n \rightarrow \infty }, we obtain

\mathrm{ \frac{x}{6}-0<\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n\left[k^2 x\right] \leq \frac{x}{6} \Rightarrow \lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n\left[k^2 x\right]=\frac{x}{6} . }

Hence option 3 is correct.






 

Posted by

avinash.dongre

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