Proof- central term of recursive pyramid 
Moderator Note: At the time that this question was posted, it was from an ongoing contest. The relevant deadline has now passed. 

How can we prove $4^{-n}(n+1)\left(2^{2n+1}-{2n+1\choose n+1}\right)$ produces  the central term in the $2n+1$th row of the following pyramid?
The pyramid is produced in this way: atop the triangle is 1 and each number below it is determined by summing half of each of the numbers that it is supporting, and adding 1 to the sum.                                         
                             1
                        3/2     3/2
                   7/4     10/4     7/4
              15/8    25/8      25/8    15/8 
        31/16     56/16    66/16    56/16   31/16

I have attempted induction and it does not work well as I am sure you can see. I do not know how to prove the formula... even an intuitive proof would work. 
Please see my comment below about constructing the pyramid as a whole using coefficients. Another question to all(ESP joriki) : what can be said in general about any term in the pyramid? Can we make a generalization of any term in the pyramid based on The row number and the $k+1$th or kth term in a row, perhaps, depending on how you define k? I have been looking at it and it seems that writing it in terms of k and z, where z=n-1 might make it simple, but I don't know. 

It appears that joriki has answered the original question. I got bogged down on that when I got sick, but I did manage to establish that $4^{-n}(n+1)\left(2^{2n+1}-{2n+1\choose n+1}\right)$ gives the expected number of tosses of a fair coin required to get either $n+1$ heads or $n+1$ tails, as I mentioned in the comments. It’s too long for a comment, but it might be of some interest, so here it is.

Suppose that we toss a fair coin until we get either $n+1$ heads or $n+1$ tails. Say that this happens on the $k$-th toss; clearly $n+1\le k\le 2n+1$. There are $\binom{k-1}n$ sequences that terminate with a head on the $k$-th toss and another $\binom{k-1}n$ sequences that terminate with a tail on the $k$-th toss. Each of these occurs with probability $2^{-k}$, so the expected number of tosses is
$$2\sum_{k=n+1}^{2n+1}2^{-k}k\binom{k-1}n\;.$$
To get rid of the negative exponents I’m going to multiply through by $2^{2n}$ to get
$$\begin{align*}
2^{2n+1}\sum_{k=n+1}^{2n+1}2^{-k}k\binom{k-1}n&=\sum_{k=n+1}^{2n+1}2^{2n+1-k}(k-n)\binom{k}n\\
&=\sum_{k=1}^{n+1}2^{n+1-k}k\binom{n+k}n\\
&=\sum_{k=1}^{n+1}2^{n+1-k}(n+k)\binom{n+k-1}{k-1}\\
&=\sum_{k=0}^n2^{n-k}(n+1+k)\binom{n+k}k\\
&=\sum_{k=0}^n2^{n-k}(n+1)\binom{n+1+k}k\\
&=(n+1)\sum_{k=0}^n2^{n-k}\binom{n+1+k}k\;.
\end{align*}$$
Thus, we’d like to show that $$\sum_{k=0}^n2^{n-k}\binom{n+1+k}k=2^{2n+1}-\binom{2n+1}{n+1}\;.\tag{1}$$
Let $$s_n=\sum_{k=0}^n2^{n-k}\binom{n+1+k}k\;.$$ Then 
$$\begin{align*}s_{n+1}&=\sum_{k=0}^{n+1}2^{n+1-k}\binom{n+2+k}k\\
&=\sum_{k=0}^{n+1}2^{n+1-k}\left(\binom{n+1+k}k+\binom{n+1+k}{k-1}\right)\\
&=2\sum_{k=0}^n2^{n-k}\binom{n+1+k}k+\binom{2n+2}{n+1}+\sum_{k=0}^n2^{n-k}\binom{n+2+k}k\\
&=2s_n+\binom{2n+2}{n+1}+\frac12\left(s_{n+1}-\binom{2n+3}{n+1}\right)\;,
\end{align*}$$
so $$s_{n+1}=4s_n+2\binom{2n+2}{n+1}-\binom{2n+3}{n+1}\;.$$ Taking $(1)$ as induction hypothesis, we have 
$$\begin{align*}
s_{n+1}&=2^{2n+3}-4\binom{2n+1}{n+1}+2\binom{2n+2}{n+1}-\binom{2n+3}{n+1}\\
&=2^{2n+3}-\binom{2n+3}{n+2}+2\left(\binom{2n+1}{n+1}+\binom{2n+1}n\right)-4\binom{2n+1}{n+1}\\
&=2^{2n+3}-\binom{2n+3}{n+2}+2\left(\binom{2n+1}{n+1}+\binom{2n+1}{n+1}\right)-4\binom{2n+1}{n+1}\\
&=2^{2n+3}-\binom{2n+3}{n+2}\;,
&=
\end{align*}$$
as desired.
