Question

If $0<a_1<a_2<....<a_n$,then prove that $ta{n}^{-1}\left(\frac{a_{1}x-y}{x+a_{1}y}\right)+ta{n}^{-1}\left(\frac{a_2-a_1}{1+a_2{a}_1}\right)+ta{n}^{-1}\left(\frac{a_3-a_2}{1+a_3{a}_2}\right)+...+ta{n}^{-1}\left(\frac{a_n-a_{n-1}}{1+a_n{a}_{n-1}}\right)+ta{n}^{-1}\left(\frac{1}{a_n}\right)$ is of the form $k_1\pi +ta{n}^{-1}\frac{k_{2}x}{y}$, then $8k_1{k}_2$ is equal to

Answer 1

Consider
$ta{n}^{-1}\left(\frac{a_n-a_{n-1}}{1+a_n.a_{n-1}}\right)$
$=ta{n}^{-1}\left(\frac{\frac{1}{a_{n-1}}-\frac{1}{a_n}}{1+\frac{1}{a_n}.\frac{1}{a_{n-1}}}\right)$
$=ta{n}^{-1}\left(\frac{1}{a_{n-1}}\right)-ta{n}^{-1}\left(\frac{1}{a_n}\right)$
Hence
$ta{n}^{-1}\left(\frac{a_2-a_1}{1+a_1.a_2}\right)+ta{n}^{-1}\left(\frac{a_3-a_2}{1+a_2.a_3}\right)+....+ta{n}^{-1}\left(\frac{1}{a_n}\right)$
$=ta{n}^{-1}\left(\frac{1}{a_1}\right)-ta{n}^{-1}\left(\frac{1}{a_2}\right)+ta{n}^{-1}\left(\frac{1}{a_2}\right)-ta{n}^{-1}\left(\frac{1}{a_3}\right)....+ta{n}^{-1}\left(\frac{1}{a_{n-1}}\right)-ta{n}^{-1}\left(\frac{1}{a_n}\right)+ta{n}^{-1}\left(\frac{1}{a_n}\right)$
$=ta{n}^{-1}\left(\frac{1}{a_1}\right)$ ...(i)
Now
$ta{n}^{-1}\frac{a_{1}x-y}{a_{1}y+x}+ta{n}^{-1}\left(\frac{1}{a_1}\right)$
$=ta{n}^{-1}\left(\frac{\frac{a_{1}x-y}{a_{1}y+x}+\frac{1}{a_1}}{1-\frac{1}{a_1}.\frac{a_{1}x-y}{a_{1}y+x}}\right)$
$=ta{n}^{-1}\left(\frac{a_1^{2}x-a_{1}y+a_{1}y+x}{a_1^{2}y+a_{1}x-a_{1}x+y}\right)$
$=ta{n}^{-1}\left(\frac{a_1^{2}x+x}{a_1^{2}y+y}\right)$
$=ta{n}^{-1}\left(\frac{x}{y}\right)$.
Hence $k_1=0$ and $k_2=1$.