An algebra problem by Department 8

My solution is same as Chew-Seong Cheong , but a different approach:-

$\large{S=\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { \left( k \right) }^{ 2 }\left( k+1 \right) { \left( k+2 \right) }^{ 2 } } } \\ =\sum _{ k=2 }^{ \infty }{ \frac { 1 }{ { \left( k-1 \right) }^{ 2 }\left( k \right) { \left( k+1 \right) }^{ 2 } } } \\ =\sum _{ k=2 }^{ \infty }{ \frac { k }{ { \left[ \left( k-1 \right) \left( k \right) \left( k+1 \right) \right] }^{ 2 } } } \\ =\frac { 1 }{ 4 } \left( \sum _{ k=2 }^{ \infty }{ \left( \frac { 4k }{ { \left[ \left( k-1 \right) \left( k \right) \left( k+1 \right) \right] }^{ 2 } } \right) } \right) \\ =\frac { 1 }{ 4 } \left( \sum _{ k=2 }^{ \infty }{ \left( \frac { 1 }{ { \left[ \left( k-1 \right) \left( k \right) \right] }^{ 2 } } -\frac { 1 }{ \left[ \left( k \right) \left( k+1 \right) \right] } \right) } \right) \\ =\frac { 1 }{ 4 } \left( \frac { 1 }{ 1\times { 2 }^{ 2 } } -\frac { 1 }{ { 2 }^{ 2 }\times { 3 }^{ 2 } } +\frac { 1 }{ { 2 }^{ 2 }\times { 3 }^{ 2 } } \cdots \cdots \cdots \right) \\ S=\frac { 1 }{ 16 } }$

Now we have

${ S }^{ -497 }={ 16 }^{ 497 }\equiv { 2 }^{ 497 }(\mod{7})$

Now by Fermat's Little theorem

${ 2 }^{ 497 }\equiv { 2 }^{ 5 }\equiv \boxed{4} (\mod{7})$

$\begin{aligned} S & = \sum_{k=1}^\infty \frac{1}{k^2(k+1)(k+2)^2} \\ & = \sum_{k=1}^\infty \left(\frac{1}{k(k+2)}\times \frac{1}{k(k+1)(k+2)}\right) \\ & = \sum_{k=1}^\infty \left[ \frac{1}{4} \left(\frac{1}{k} - \frac{1}{k+2} \right) \left(\frac{1}{k} - \frac{2}{k+1} + \frac{1}{k+2} \right) \right] \\ & = \frac{1}{4} \sum_{k=1}^\infty \left(\frac{1}{k^2} - \frac{2}{k(k+1)} + \frac{1}{k(k+2)} - \frac{1}{k(k+2)} + \frac{2}{(k+1)(k+2)} - \frac{1}{(k+2)^2} \right) \\ & = \frac{1}{4} \sum_{k=1}^\infty \left(\frac{1}{k^2} - \frac{1}{(k+2)^2} - 2 \left[ \frac{1}{k(k+1)} - \frac{1}{(k+1)(k+2)} \right] \right) \\ & = \frac{1}{4} \left( \sum_{k=1}^\infty \frac{1}{k^2} - \sum_{k=1}^\infty \frac{1}{(k+2)^2} - 2 \left[ \sum_{k=1}^\infty \frac{1}{k(k+1)} - \sum_{k=1}^\infty \frac{1}{(k+1)(k+2)} \right] \right) \\ & = \frac{1}{4} \left( \sum_{k=1}^\infty \frac{1}{k^2} - \sum_{k=3}^\infty \frac{1}{k^2} - 2 \left[ \sum_{k=1}^\infty \frac{1}{k(k+1)} - \sum_{k=2}^\infty \frac{1}{k(k+1)} \right] \right) \\ & = \frac{1}{4} \left( \frac{1}{1} + \frac{1}{4} - 2 \left[ \frac{1}{2} \right] \right) \\ & = \frac{1}{16} \end{aligned}$

$\begin{aligned} \\ & \\ \Rightarrow S^{-497} & \equiv 16^{497} \equiv 2^{1988} \equiv 2^28^{662} \equiv 4(7+1)^{662} \equiv \boxed{4} \pmod{7} \end{aligned}$