This question was uploaded on 29/01/22 on social media accounts.
Question from JEE Advanced question paper.
Solution:
Consider 8 as 23 in all places
We know that:
\[\color{black} {a^3-b^3=(a-b)(a^2+ab+b^2)}\]
\[\color{black} {a^3+b^3=(a+b)(a^2-ab+b^2)}\]
Now,
\[\color{black} {\frac{n^3-2^2}{n^3+2^3}=\color{blue}{\left(\frac{n-2}{n+2}\right)}\color{red}{\left(\frac{n^2+2n+4}{n^2-2n+4}\right)}}\]
Solving in two parts:
Let's consider the blue part from all terms in the series:
\[\color{blue}{\left(\frac{3-2}{3+2}\right)\left(\frac{4-2}{4+2}\right)\left(\frac{5-2}{5+2}\right)....=\frac{1\cdot2\cdot3\cdot4\cdot5\cdot\cdot\cdot\cdot}{5\cdot6\cdot7\cdot8\cdot9\cdot\cdot\cdot\cdot}=24}\]
Now consider the red part from all terms in the series:
\[\color{red}{\frac{19\cdot28\cdot39\cdot52\cdot67\cdot\cdot\cdot\cdot}{7\cdot12\cdot19\cdot28\cdot39\cdot\cdot\cdot\cdot}=\frac{1}{7\cdot12}}\]
Now by combining both parts:
\[\color{black} {S=\left(\color{blue}{24}\right)\left(\color{red}{\frac{1}{7\cdot12}}\right)=\frac27}\]
