Product of nested Radicals with different indices asked in the Brilliant.org

Exercise:

Show that, $\prod\limits_{i=1}^{8}{\sqrt[i+2]{\sqrt[i+1]{32}}=4}$

Solution: we know that $\sqrt[m]{{{a}^{n}}}={{a}^{\frac{n}{m}}}$ and ${{\left( {{a}^{n}} \right)}^{t}}={{a}^{nt}}$

So it’s clear that $\sqrt[i+2]{\sqrt[i+1]{32}}={{\left( {{32}^{\frac{1}{i+1}}} \right)}^{\frac{1}{i+2}}}={{32}^{\frac{1}{i+1}\times \frac{1}{i+2}}}$

Hence $\prod\limits_{i=1}^{8}{\sqrt[i+2]{\sqrt[i+1]{32}}}={{32}^{\sum\limits_{i=1}^{8}{\frac{1}{i+1}\times \frac{1}{i+2}}}}$ (since ${{a}^{n}}\times {{a}^{m}}={{a}^{n+m}}$ )

So Let’s compute that sum .

$\sum\limits_{i=1}^{8}{\frac{1}{\left( i+1 \right)\left( i+2 \right)}=\sum\limits_{i=1}^{8}{\left( \frac{A}{i+1}+\frac{B}{i+2} \right)}}=\sum\limits_{i=1}^{8}{\frac{Ai+2A+Bi+B}{\left( i+1 \right)\left( i+2 \right)}=\sum\limits_{i=1}^{8}{\frac{i\left( A+B \right)+2A+B}{\left( i+1 \right)\left( i+2 \right)}}}$

Hence $A+B=0\,\,\And \,\,2A+B=0$ then $A=1\,\And \,B=-1$

Thus $\sum\limits_{i=1}^{8}{\frac{1}{\left( i+1 \right)\left( i+2 \right)}=\sum\limits_{i=1}^{8}{\frac{1}{i+1}-\sum\limits_{i=1}^{8}{\frac{1}{i+2}}}}$

$=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{9}-\left( \frac{1}{3}+\frac{1}{4}+....+\frac{1}{9}+\frac{1}{10} \right)$

$=\frac{1}{2}+\left( \frac{1}{3}-\frac{1}{3} \right)+\left( \frac{1}{4}-\frac{1}{4} \right)+...+\left( \frac{1}{9}-\frac{1}{9} \right)-\frac{1}{10}$

$=\frac{1}{2}-\frac{1}{10}=\frac{5-1}{10}=\frac{4}{10}=\frac{2}{5}$  but $32={{2}^{5}}$ thus ${{32}^{\sum\limits_{i=1}^{8}{\frac{1}{\left( i+1 \right)\left( i+2 \right)}}}}={{32}^{2/5}}={{\left( {{2}^{5}} \right)}^{2/5}}={{2}^{2}}=4$