Compare the coefficients

Since we are looking for the coefficient of a term with even power (viz $x^{800}$ ) in a polynomial, its value remain unchanged if we replace $x$ by $-x$ in the expressions. Thus $C_1$ and $C_2$ are also the coefficients of $x^{800}$ in the polynomials $Q_1(x)=P_1(-x)=(1+x^2+x^5)^{2000}, \hspace{5pt}Q_2(x)=P_2(-x)=(1-x^2-x^5)^{2000}$ It is clear that all coefficients of the polynomial $Q_1(x)$ are positive and coefficient of $x^{800}$ is non-zero. Whereas some of the terms contributing to $x^{800}$ in $Q_2(x)$ are negative. Thus the result follows.

To make 800 in the power of x,in the first polynomial ,we would require even number of $x^5$ .Since, $x^2$ has coefficient positive,if we assume we have multiplied the whole thing,every contribution to $x^{800}$ would be positive. But,in the second polynomial,we can have negative contribution to $x^{800}$ ( for example multiply 158 $x^5$ and 5 $x^2$ .Since the absolute value of all contribution are same for both the polynomial, $C_1 > C_2$

let $a$ be the times choosing $x^2$ , and $b$ the times choosing $x^5$ . In order to make term $x$ to the 800, we would have the equation: $2a+5b=800$

We see that $b$ must be even, meaning that for the first equation the terms adding to the coefficient are all positive. For $a$ , it must be divisible by 5, but can be even or odd, making the coefficient in second equation contain both positive terms and negative terms. Because both the equations have the same magnitude of terms, only differing in signs, the coefficient of x to the 800 in the first equation, added only by positive terms, must be bigger than that in the second equation, which is made of both positive and negative terms.

Thus, the answer is B

In order to form $x^{800}$ , the number of $x^5$ 's contributing to the term would have to be even; therefore, all the $x^{800}$ terms in $P_1$ , e.g. $(x^2)^{15}(-x^5)^{10}$ , would be positive. On the other hand, the number of $x^2$ 's contributing to form $x^{800}$ would have to be a multiple of five, but because this could be an odd multiple of five, some of the $x^{800}$ terms in $P_2$ , e.g. $(-x^2)^{15}(x^5)^{10}$ , would be negative. Thus $C_2$ would have to be greater than $C_1$ .