Evaluating a summation

Let us try to generalise it:

Consider $(1-x)^{n} = \displaystyle \sum_{k=0}^{n} (-1)^{k} {n \choose k} x^k$

$\Rightarrow x(1-x)^n = \displaystyle \sum_{k = 1}^{n+1} (-1)^{k-1} {n \choose k-1} x^k$

Integrate both sides in 0 to 1 to get :

$\displaystyle \sum_{k=1}^{n+1} \frac{(-1)^{k-1} {n \choose k-1}}{k+1} = \int_{0}^{1} x(1-x)^n \text{dx}$

= $\displaystyle \int_{0}^{1} x^{n}(1-x) \text{dx}$ (Since, $\displaystyle \int_{0}^{a} f(x) \text{dx} = \int_{0}^{a} f(a-x) \text{dx}$ )

= $\displaystyle \int_{0}^{1} x^{n} \text{dx} - \int_{0}^{1} x^{n+1} \text{dx}$

= $\boxed{\frac{1}{(n+1)(n+2)}}$

Hence, if $n=100$ , the value comes $\frac{1}{10302}$ , implying $a+b = 10303$

Since $100 \equiv 0 \pmod{2}$ , $100-(k-1)$ and $k-1$ have the same parity, so $(-1)^{100-(k-1)}= (-1)^{k-1}$ Let's ignore the $k+1$ terms in the denominator for a while. Consider the numerator of a general term: $(-1)^{k-1} \binom{100}{k-1}= (-1)^{100-(k-1)} \binom{100}{k-1}$ Notice that this is simply the coefficient of $x^{k-1}$ in $(x-1)^{100}$ , which is also equal to the coefficient of $x^k$ in $x(x-1)^{100}$ . We then conclude $x(x-1)^{100}= \sum \limits_{k=1}^{101} (-1)^{k-1} \binom{100}{k-1} x^k$ Integrating both sides from $0$ to $1$ , $\int \limits_{0}^{1} \sum \limits_{k=1}^{101} (-1)^{k-1} \binom{100}{k-1} x^k dx = \int \limits_{0}^{1} x(x-1)^{100} dx$ $\implies \left [ \sum \limits_{k=1}^{101} \frac{(-1)^{k-1} \binom{100}{k-1}}{k+1} x^{k+1} \right ]_{0}^{1}= \int \limits_{0}^{1} x(x-1)^{100} dx$ $\implies \sum \limits_{k=1}^{101} \frac{(-1)^{k-1} \binom{100}{k-1}}{k+1}= \int \limits_{0}^{1} x(x-1)^{100} dx$ To compute the integral in the RHS, we make the substitution $y=x-1$ , so $dy=dx$ and $\begin{array}{lcl} \int \limits_{0}^{1} x(x-1)^{100} dx & = & \int \limits_{0}^{1} (y+1)y^{100} dy \\ & = & \int \limits_{0}^{1} y^{101}+y^{100} dy \\ & = & \left [ \frac{y^{102}}{102} + \frac{y^{101}}{101} \right ] _{0}^{1} \\ &= & \left [ \frac{(x-1)^{102}}{102} + \frac{(x-1)^{101}}{101} \right ] _{0}^{1}\\ & = & \frac{1}{101}-\frac{1}{102} \\ &= & \frac{1}{10302} \end{array}$ Hence, $a=1$ , $b=10302$ , and $a+b= 10303 \equiv \boxed{303} \pmod{1000}$ .