Let S_{k}=_{r=1}^{k} ^{-1}≤ft({6^{r}}{2²r+1}+3²r+1}}). Then, _{k→∞} S_{k} is equal to [16 Mar 2021 Shift 1]

Correct Answer is : cot⁡−1(32)^{-1}≤ft(3/2)cot−1(23)

Correct Answer is : cot⁡−1(32)^{-1}≤ft(3/2)cot−1(23)

Test Index

Limits, Continuity and Differentiability Part 2

Section: Mathematics

Question : 88 of 100

Marks: +1, -0

Let

S_{k}=\sum\limits_{r=1}^{k} \tan^{-1}\left(\frac{6^{r}}{2^{2r+1}+3^{2r+1}}\right)

\lim\limits_{k\to\infty} S_{k}

Solution:

S_{K}=\sum\limits_{r=1}^{k} \tan^{-1}\left(\frac{6^{r}}{2^{2r+1}+3^{2r+1}}\right)

=\sum\tan^{-1}\frac{2^{r}3^{r+1}-3^{r}2^{r+1}}{2^{2r+1}\left[1+\left(\frac{3}{2}\right)^{2r+1}\right]}

=\sum\tan^{-1}\frac{2^{2r+1}\left[\left(\frac{3}{2}\right)^{r+1}-\left(\frac{3}{2}\right)^{r}\right]}{2^{2t+1}\left[1+\left(\frac{3}{2}\right)^{2r+1}\right]}

=\sum\limits_{r=1}^{k} \tan^{-1}\frac{\left(\frac{3}{2}\right)^{r+1}-\left(\frac{3}{2}\right)^{r}}{1+\left(\frac{3}{2}\right)^{2r+1}}

=\sum\limits_{r=1}^{k} \tan^{-1}\frac{\left(\frac{3}{2}\right)^{r+1}-\left(\frac{3}{2}\right)^{r}}{1+\left(\frac{3}{2}\right)^{2r+1}}

=\sum\limits_{r=1}^{k} \tan^{-1}\left(\frac{3}{2}\right)^{r+1}-\tan^{-1}\left(\frac{3}{2}\right)^{r}

=\tan^{-1}\left(\frac{3}{2}\right)^{2}-\tan^{-1}\left(\frac{3}{2}\right)

+\tan^{-1}\left(\frac{3}{2}\right)^{3}-\tan^{-1}\left(\frac{3}{2}\right)^{2}

+\tan^{-1}\left(\frac{3}{2}\right)^{k+1}-\tan^{-1}\left(\frac{3}{2}\right)^{k}

\Rightarrow \quad S_{k}=\tan^{-1}\left(\frac{3}{2}\right)^{k+1}-\tan^{-1}\left(\frac{3}{2}\right)

When

k\to\infty,\quad \tan^{-1}\left(\frac{3}{2}\right)^{k+1}\to\frac{\pi}{2}

\lim\limits_{k\to\infty}s_{k}=\frac{\pi}{2}-\tan^{-1}\left(\frac{3}{2}\right)=\cot^{-1}\left(\frac{3}{2}\right)

Go to Question:

Prev Question

Next Question