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  • If the rth term  of a series is given by <img alt="t_r=\frac{r}{1+r^2+r^4}" src="https://entranceco

If the rth term t_{r} of a series is given by t_r=\frac{r}{1+r^2+r^4} find the value of \lim _{n \rightarrow \infty} \sum_{r=1}^n t_r

 

Option: 1

1


Option: 2

2


Option: 3

0


Option: 4

1/2


Answers (1)

best_answer

We have , 

\begin{aligned} t_r & =\frac{r}{1+r^2+r^4}=\frac{1}{2}\left(\frac{2 r}{\left(1+r^2+r\right)\left(1+r^2-r\right)}\right) \\ \\& =\frac{1}{2}\left(\frac{1}{\left(1+r^2+r\right)}-\frac{1}{\left(1+r^2-r\right)}\right) \end{aligned}

Thus, 

\begin{aligned} & t_r=\frac{1}{2}\left(\frac{1}{1+r(r-1)}-\frac{1}{1+r(r+1)}\right) \\ \therefore \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: & t_1=\frac{1}{2}\left(1-\frac{1}{3}\right) \\ & t_2=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{7}\right) \end{aligned}

                     t_3=\frac{1}{2}\left(\frac{1}{7}-\frac{1}{13}\right)

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\begin{aligned} t_n & =\frac{1}{2}\left(\frac{1}{1+n(n-1)}-\frac{1}{1+n(n+1)}\right) \\ \\\therefore \quad & \lim _{n \rightarrow \infty} \sum_{r=1}^n t_r=\lim _{n \rightarrow \infty}\left(1-\frac{1}{1+n(n+1)}\right) \\ \\& =\frac{1}{2}\left(1-\frac{1}{\infty}\right)=\frac{1}{2} . \end{aligned}

Posted by

Nehul

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