Dirty Polynomial (200 followers problem)

We have $P\left( x \right) ={ a }_{ 1 }{ x }^{ 200 }+{ a }_{ 2 }{ x }^{ 199 }+{ a }_{ 3 }{ x }^{ 198 }+{ a }_{ 4 }{ x }^{ 197 }+{ a }_{ 5 }{ x }^{ 196 }+{ a }_{ 6 }{ x }^{ 195 }\dots +{ a }_{ 201 }$

Let the sum of coefficients of even degree terms be A.

$A={ a }_{ 1 }+{ a }_{ 3 }+{ a }_{ 5 }+{ a }_{ 7 }+{ a }_{ 9 }+{ a }_{ 11 }+{ a }_{ 13 }+{ a }_{ 15 }+\dots +{ a }_{ 201 }$

Now finding the values of P(1) and P(-1) $P\left( 1 \right) ={ a }_{ 1 }{ 1 }^{ 200 }+{ a }_{ 2 }{ 1 }^{ 199 }+{ a }_{ 3 }{ 1 }^{ 198 }+{ a }_{ 4 }1^{ 197 }+{ a }_{ 5 }{ 1 }^{ 196 }+{ a }_{ 6 }{ 1 }^{ 195 }\dots +{ a }_{ 201 }\\ \qquad ={ a }_{ 1 }+{ a }_{ 2 }+{ a }_{ 3 }+{ a }_{ 4 }+{ a }_{ 5 }+{ a }_{ 6 }+{ a }_{ 7 }+{ a }_{ 8 }+\dots +{ a }_{ 201 }$

$P\left( -1 \right) ={ a }_{ 1 }{ \left( -1 \right) }^{ 200 }+{ a }_{ 2 }{ \left( -1 \right) }^{ 199 }+{ a }_{ 3 }{ \left( -1 \right) }^{ 198 }+{ a }_{ 4 }\left( -1 \right) ^{ 197 }+{ a }_{ 5 }{ \left( -1 \right) }^{ 196 }+{ a }_{ 6 }\left( -1 \right) ^{ 195 }\dots +{ a }_{ 201 }\\ \qquad ={ a }_{ 1 }-{ a }_{ 2 }+{ a }_{ 3 }-{ a }_{ 4 }+{ a }_{ 5 }-{ a }_{ 6 }+{ a }_{ 7 }-{ a }_{ 8 }+\dots +{ a }_{ 201 }$

By these equations, $P\left( -1 \right) +P\left( 1 \right) =2\left( { a }_{ 1 }+{ a }_{ 3 }+{ a }_{ 5 }+{ a }_{ 7 }+{ a }_{ 9 }+{ a }_{ 11 }+{ a }_{ 13 }+{ a }_{ 15 }+\dots +{ a }_{ 201 } \right)$

This gives us $P\left( -1 \right) +P\left( 1 \right) =2A$ .

So $A=\frac { P\left( -1 \right) +P\left( 1 \right) }{ 2 }$

Now putting the value of P(x) for integer x in domain -20 to 20, we get

$A=\frac { \frac { -1-2 }{ { \left( -1 \right) }^{ 2 }+1 } +\frac { 1-2 }{ { 1 }^{ 2 }+1 } }{ 2 }$

Simplifying, $A=\boxed { -1 }$