Reducing Binomial Summation

Question

How can I reduce this summation into this $$\frac{1}{2}(1+\left(1/3\right)^{50})$$ The problem comes from the 1992 AHSME Test (problem 29)

Answer 1

While there are definitely shorter solutions, here is my attempt to be thorough. Have you heard of the binomial distribution in probability theory?

Say we have a random variable $X$ that follows the binomial distribution. For this purpose, I will say that a random variable is a variable that can take on certain possible values from a set. Thus, $x\sim Bin(n,p)$ where $n$ is the number of trials and $p$ is the probability of success. The probability of getting $k$ successes in $n$ trials $($think of fair coin tosses with $n$ trials and $p=\frac{1}{2}$$)$ is given by

$$P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$$

$X$ can take on values from $\{0,1,\cdots,n\}$, quite reasonably.

By the axiom of total probability, $$\sum_{k=0}^nP(X=k)=\sum_{k=0}^n\binom{n}{k}p^k(1-p)^{n-k}=1$$

Thus, we have something similar here but only the even terms:

$$\sum_{n=0}^{25} \binom{50}{2n} (2/3)^{2n}(1/3)^{50-2n}=\sum_{n=0}^{50}\binom{50}{n}(2/3)^n(1/3)^{50-n}-\sum_{n=1}^{25}\binom{50}{2n-1}(2/3)^{2n-1}(1/3)^{50-2n+1}$$

Upon close examination, it becomes obvious that the difference between the even terms and the odd terms is just the first term: $$\sum_{n=0}^{25} \binom{50}{2n} (2/3)^{2n}(1/3)^{50-2n}-\sum_{n=1}^{25}\binom{50}{2n-1}(2/3)^{2n-1}(1/3)^{50-2n+1}=\frac{1}{3^{50}}$$

Thus, we have \begin{align} x+y&=1\\ x-y&=\frac{1}{3^{50}} \end{align}

This leads to the desired result $$\frac{1}{2}\left(1+\frac{1}{3^{50}}\right)$$