While working on probabalistic question and binomial distribution got the following result:
$$\sum_{k=0}^{n} \binom{n}{k}\sum_{l=0}^{k}\binom{n+1}l = 2^{2n}$$
or in other words sum of binomial coefficients multiplied by the partial sum of binomial sequence of $n+1$ order.
Does it have some nice explanation?
The sum is $$ \begin{align} S&=\sum_{k=0}^n\binom{n}k\sum_{l=0}^k\left[\binom{n}{l-1}+\binom{n}{l}\right]\\ &=\sum_{k=0}^n\binom{n}k\left[\binom{n}k+2\sum_{l=0}^{k-1}\binom nl\right]\\ &=\sum_{k=0}^n\binom{n}{k}^2+2\sum_{j,k\colon 0\le l<k\le n} \binom nl\binom nk. \end{align}$$ The double sum here is the sum of $\binom nl\binom nk$ over all $k$, $l\in\{0,\ldots,n\}$ with $k\ne l$. Thus $$S=\sum_{k=0}^n\sum_{l=0}^n\binom nl\binom nk=(1+1)^n(1+1)^n.$$