Counting, Probability and Binomial Coefficients If $$P_{2n+2}=\sum_{k=n+2}^{2n+2}{2n+2 \choose k}p^kq^{2n+2-k}$$
and,
$$P_{2n}=\sum_{k=n+1}^{2n}{2n \choose k}p^kq^{2n-k}$$
where $0<p<q<1$ and $q=1-p$
Prove that 
$$P_{2n+2}=P_{2n}+{2n \choose n}p^{n+2}q^n-{2n \choose {n+1}}p^{n+1}q^{n+1}$$
$\mathbf {Inspiration:}$ A and B play a series of games where the probability of winning $\mathit p$ for A is kept less than 0.5. However A gets to choose in advance the total no. of plays. To win the game one must score more than half the games . If the total no. of games is to be even, How many plays should A choose?
$\mathbf {Here}$ $P_{2n}$ and $P_{2n+2}$ represents the probability of A winning the play in $2n$ and $2n+2$ games where $2n$ is considered the optimum number of games

We consider $(p+q)^2P_{2n}$ instead of $P_{2n}$ (because then our identity will be homogeneous). Then, we will simply compare coefficents of monomials $p^kq^{2n+2-k}$ in both sides of identity.
Note that (from equality $p+q=1$)
$$
P_{2n}=\sum_{k=n+1}^{2n}{2n \choose k}p^kq^{2n-k}=\sum_{k=n+1}^{2n}{2n \choose k}p^kq^{2n-k} (p+q)^2,
$$
which equals
$$
\sum_{k=n+1}^{2n}{2n \choose k}p^kq^{2n-k} (p^2+2pq+q^2)=
\\
\sum_{k=n+1}^{2n}{2n \choose k}p^{k+2}q^{2n-k}+\sum_{k=n+1}^{2n}2{2n \choose k}p^{k+1}q^{2n-k+1}+\sum_{k=n+1}^{2n}{2n \choose k}p^{k}q^{2n-k+2}=
\\
\sum_{k=n+3}^{2n+2}{2n \choose k-2}p^{k}q^{2n+2-k}+\sum_{k=n+2}^{2n+1}2{2n \choose k-1}p^{k}q^{2n+2-k}+\sum_{k=n+1}^{2n}{2n \choose k}p^{k}q^{2n-k+2}.
$$
Therefore, $P_{2n}$ equals
$$
\sum_{k=n+2}^{2n+2}\left({2n \choose k}+2{2n \choose k-1}+{2n \choose k-2}\right)p^kq^{2n+2-k}-\left({2n \choose n}p^{n+2}q^{n}-{2n \choose n+1}p^{n+1}q^{n+1}\right).
$$
We also know that 
$$
{2n \choose k}+2{2n \choose k-1}+{2n \choose k-2}=\left({2n \choose k}+{2n \choose k-1}\right)+\left({2n \choose k-1}+{2n \choose k-2}\right)=
\\
{2n+1 \choose k}+{2n+1 \choose k-1}={2n+2 \choose k},
$$
so we obtain
$$
P_{2n}=\sum_{k=n+2}^{2n+2}{2n+2 \choose k}p^kq^{2n+2-k}=P_{2n+2}-\left({2n \choose n}p^{n+2}q^{n}-{2n \choose n+1}p^{n+1}q^{n+1}\right).
$$
Hence,
$$
P_{2n+2}=P_{2n}+{2n \choose n}p^{n+2}q^{n}-{2n \choose n+1}p^{n+1}q^{n+1},
$$
as desired.
