Calculate the summations
Hello, WORLD!
This is a tough problem for the algebra forum. I'm going to use a little calculus. I hope you will still find my remarks useful.
Start with the Binomial Theorem:
$\displaystyle (1+x)^n = \sum_{i=0}^n \binom{n}{i} x^i$
Integrate both sides from 0 to x (this is the calculus part). The result is
$\displaystyle \frac{1}{n+1}[(1+x)^{n+1}-1] = \sum_{i=0}^n \frac{1}{i+1} \binom{n}{i} x^{i+1}$
Now multiply these two equations together. I'm going to replace i with j in the first equation, so it will be easier to keep everything straight.
(*)....$\displaystyle (1+x)^n \cdot \frac{1}{n+1}[(1+x)^{n+1}-1] = \sum_{j=0}^n \binom{n}{j} x^j \cdot \sum_{i=0}^n \frac{1}{i+1} \binom{n}{i} x^{i+1}$
Now consider the coefficient of $\displaystyle x^{n+1}$ in (*). The left-hand side of (*) is equal to
$\displaystyle \frac{1}{n+1}[(1+x)^{2n+1} - (1+x)^{n}] = \frac{1}{n+1} \left[ \sum_{i=0}^{2n+1} \binom{2n+1}{i} x^i - \sum_{i=0}^n \binom{n}{i}x^i \right] $
so the coefficient of $\displaystyle x^{n+1}$ is
$\displaystyle \frac{1}{n+1} \binom{2n+1}{n+1}$
On the right-hand side of (*), the coefficient of $\displaystyle x^{n+1}$ occurs when j = n-i, so the coefficient is
$\displaystyle \sum_{i=0}^n \binom{n}{n-i} \cdot \frac{1}{i+1} \binom{n}{i} = \sum_{i=0}^n \frac{1}{i+1} \binom{n}{i}^2$
since $\displaystyle \binom{n}{i} = \binom{n}{n-i}$.
The coefficient of $\displaystyle x^{n+1}$ must be the same on both sides of the equation, so we have
$\displaystyle \frac{1}{n+1} \binom{2n+1}{n+1} = \sum_{i=0}^n \frac{1}{i+1} \binom{n}{i}^2$
Now set n = 100, and you're done.