If f(n) = [^{-1}1/1+2+^{-1}1/1+6+^{-1}1/1+12+.....+^{-1}1/1+n(n+1)] then f(2021) =

Correct Answer is : 202120232021/202320232021

Correct Answer is : 202120232021/202320232021

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TS EAMCET 5-Aug-2021 Shift 2 Question Paper

Section: Mathematics

Question : 25 of 160

Marks: +1, -0

f(n)

\tan[\tan^{-1}\frac{1}{1+2}+\tan^{-1}\frac{1}{1+6}+\tan^{-1}\frac{1}{1+12}+.....+\tan^{-1}\frac{1}{1+n(n+1)}]

f(2021)

Solution: 👈: Video Solution

f(n)=\tan[\tan^{-1}\;\frac{1}{1+2}+\tan^{-1}\;\frac{1}{1+6}+\tan^{-1}\;\frac{1}{1+12}

+.....+\tan^{-1}\;\frac{1}{1+n(n+1)}]

f(n)=\tan\left[\sum\limits_{r=1}^{n}\tan^{-1}\;\frac{1}{1+r(r+1)}\right]

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

=\tan\left[\sum\limits_{r=1}^{n}\tan^{-1}(r+1)-\tan^{-1}(r)\right]

=\tan\begin{pmatrix}\tan^{-1}2-\tan^{-1}1\\tan^{-1}3-\tan^{-1}2\\tan^{-1}4-\tan^{-1}3\\cdot s\ \cdot s\\tan^{-1}(n+1)-\tan^{-1}n\end{pmatrix}

=\tan(\tan^{-1}(n+1)-\tan^{-1}1)

=\tan\tan^{-1}\left[\frac{(n+1)-1}{1+(n+1)}\right]

=\tan\tan^{-1}\left(\frac{n}{n+2}\right)=\;\frac{n}{n+2}

n=2021 \Rightarrow \;\frac{2021}{2023}=f(2021)

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