Partial sum of the sequence of combinations of 256

Question

Consider the following sequence:

$A_n = \frac{256 P_n}{n!}$

Where $256 P_n = \frac {256!}{(256-n)!} $ and $0 < n \leq 128$.

Is there a closed form formula for the partial sum of this sequence, and if there how to calculate it?

Answer 1

Not very useful, I guess.

As Lord Shark the Unknown commented, "there is no nice formula for general partial sums of the rows of Pascal's triangle".

However, there is one which is quite complex $$S_p=\sum_{n=0}^{p}\frac{256!}{n!(256-n)!} =\sum_{n=0}^{p}\binom{256}n=2^{256}-\binom{256}{p+1} \, _2F_1(1,p-255;p+2;-1)$$ where appears the Gaussian or ordinary hypergeometric function.

Answer 2

I presume you want $$S=\sum_{n=0}^{128}\frac{256!}{n!(256-n)!} =\sum_{n=0}^{128}\binom{256}n.$$ This is the sum of the first half of a row of Pascal's triangle, including the middle term. The sum of a whole row of Pascal's triangle is a power of two $$\sum_{n=0}^{256}\binom{256}n=2^{256}.$$ The rows of Pascal's triangle are symmetric: $$\binom{256}n=\binom{256}{256-n}$$ so that $$S=\sum_{n=128}^{256}\binom{256}n.$$ Then $$2S=\sum_{n=0}^{128}\binom{256}n+\sum_{n=128}^{256}\binom{256}n =\sum_{n=0}^{256}\binom{256}n+\binom{256}{128} =2^{256}+\binom{256}{128}.$$