Question

If $y=ta{n}^{-1}\left(\frac{1}{1+x+x^2}\right)+ta{n}^{-1}\left(\frac{1}{x^2+3x+3}\right)+ta{n}^{-1}\left(\frac{1}{x^2+5x+7}\right)+....+$ up to n terms. Then y’(0) is equal to

• A. $-\frac{1}{1+n^2}$
• B. $-\frac{n^2}{1+n^2}$
• C. $\frac{n}{1+n^2}$
• D. $noneofthese$

Answer 1

The correct option is B $-\frac{n^2}{1+n^2}$

$y=ta{n}^{-1}\left\{\frac{1}{1+x(1+x)}\right\}+ta{n}^{-1}\left\{\frac{1}{1+(x+1)(x+2)}\right\}+ta{n}^{-1}\left\{\frac{1}{1+(x+2)(x+3)}\right\}+...+ta{n}^{-1}\left\{\frac{1}{1+(x+n-1)(x+n)}\right\}={\sum }_{r=1}^{n}ta{n}^{-1}\left\{\frac{1}{1+(x+r-1)(x+r)}\right\}={\sum }_{r=1}^{n}ta{n}^{-1}\left\{\frac{(x+r)-(x+r-1)}{1+(x+r-1)(x-r)}\right\}={\sum }_{r=1}^n\left\{ta{n}^{-1}\right(x+r)-ta{n}^{-1}(x+r-1\left)\right\}=ta{n}^{-1}(x+n)-ta{n}^{-1}x{y}^{\prime }=\frac{1}{1+(x+n{)}^2}-\frac{1}{1+x^2}\Rightarrow y^{\prime }\left(0\right)=\frac{1}{1+n^2}-1=\frac{-n^2}{1+n^2}$