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
 A: 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.
A: The equality reduces to
$$
\binom{2 n}{n+1} p^{n+1} (1-p)^{n-1} \, _2F_1\left(1,1-n;n+2;\frac{p}{p-1}\right)+\binom{2 n}{n} p^{n+2} (1-p)^n-\binom{2 n}{n+1} p^{n+1} (1-p)^{n+1}=\binom{2 (n+1)}{n+2} (1-p)^{-n+2 (n+1)-2} p^{n+2} \, _2F_1\left(1,-n;n+3;\frac{p}{p-1}\right)
$$
Let $w=p/(1-p)$, then the above is
$$
\binom{2 n}{n+1} \left((w+1)^2 \, _2F_1(1,1-n;n+2;-w)-1\right)+w \binom{2 n}{n}=w \binom{2 (n+1)}{n+2} \, _2F_1(1,-n;n+3;-w)
$$
We can then extract the coefficient for $w^m$ from both side for $m \in \{0,\dots,n\}$ to see that this equality holds.
A: We have for the first sum
$$S = \sum_{k=0}^n {2n+2\choose k+n+2} p^{k+n+2} q^{n-k}$$
and for the second one
$$T = \sum_{k=0}^{n-1} {2n\choose k+n+1} p^{k+n+1} q^{n-1-k}$$
and seek to show
$$S-T =
{2n\choose n} p^{n+2} q^n - {2n\choose n+1} p^{n+1} q^{n+1}$$
where $p+q=1.$ We see that with
$$Q_n = \sum_{k=0}^n {2n+2\choose k+n+2} p^{k+n+2} q^{n-k}$$
the claim becomes
$$Q_n - Q_{n-1} =
{2n\choose n} p^{n+2} q^n - {2n\choose n+1} p^{n+1} q^{n+1}.$$
Now
$$Q_n = p^{n+2} q^n
\sum_{k=0}^n {2n+2\choose k+n+2} p^{k} q^{-k}
= p^{n+2} q^n
\sum_{k=0}^n {2n+2\choose n-k} p^{k} q^{-k}
\\ = p^{n+2} q^n
\sum_{k=0}^n  p^{k} q^{-k}
[z^{n-k}] (1+z)^{2n+2}
= p^{n+2} q^n  [z^n] (1+z)^{2n+2}
\sum_{k=0}^n  p^{k} q^{-k} z^k.$$
We may extend $k$ beyond $n$ owing to the coefficient extrator $[z^n]$
in front:
$$p^{n+2} q^n  [z^n] (1+z)^{2n+2}
\sum_{k\ge 0}  p^{k} q^{-k} z^k
= p^{n+2} q^n  [z^n] (1+z)^{2n+2} \frac{1}{1-pz/q}
\\ = p^2  [z^n] (1+pqz)^{2n+2} \frac{1}{1-p^2z}.$$
We thus have
$$Q_{n-1} = p^2  [z^{n-1}] (1+pqz)^{2n} \frac{1}{1-p^2z}
\\ = p^2  [z^n] z (1+pqz)^{2n} \frac{1}{1-p^2z}.$$
Subtracting we find
$$Q_n - Q_{n-1}
= p^2  [z^n] ((1+pqz)^2-z) (1+pqz)^{2n} \frac{1}{1-p^2z}
\\ = p^2  [z^n] (1-p^2 z) (1-q^2 z)
(1+pqz)^{2n} \frac{1}{1-p^2z}
\\ = p^2  [z^n] (1-q^2 z) (1+pqz)^{2n}
\\ = p^2  [z^n]  (1+pqz)^{2n}
- p^2  [z^n] q^2 z (1+pqz)^{2n}
\\ = p^2 p^n q^n {2n\choose n}
- p^2 q^2 [z^{n-1}] (1+pqz)^{2n}
\\ = p^{n+2} q^n {2n\choose n}
- p^2  q^2 p^{n-1} q^{n-1} {2n\choose n-1}.$$
This is indeed
$$\bbox[5px,border:2px solid #00A000]{
p^{n+2} q^n {2n\choose n}
- p^{n+1} q^{n+1} {2n\choose n+1}}$$
as claimed.
Remark. It  might be simpler not  to merge the $p^n  q^n$ into the
coefficient extractor. Further commentary TBA.
