This question was uploaded on 05/11/21 on social media accounts.
This question was asked in Putnam 1977 B1 and I got this question from YouTube from a channel named "Letsthinkcritically"
Solution:
Let:
\[S = \frac{2^3-1}{2^3+1}\times\frac{3^3-1}{3^3+1}\times\frac{4^3-1}{4^3+1}\times\cdot\cdot\cdot\]
We know that:
\[a^3 - b^3 = (a - b) (a^2 + ab + b^2)\]
\[a^3 + b^3 = (a + b) (a^2 - ab + b^2)\]
By putting this in our original series, we will get:
\[S = \frac{1\times7}{3\times3}\times\frac{2\times13}{4\times7}\times\frac{3\times21}{5\times13}\times\cdot\cdot\cdot\]
\[\Rightarrow S = \left(\frac23\times\frac34\times\frac45\times\cdot\cdot\cdot\right)\left(\frac73\times\frac{13}7\times\frac{21}{13}\times\cdot\cdot\cdot\right)\]
\[\Rightarrow S = (2)\left(\frac13\right)=\frac23\]
