Is there any Simplification to the sum

Question

Can we simplify the following sum $$\sum_{k=0}^{10}{20\choose k}(1/4)^k (3/4)^{20-k}$$ MY TRY: I tried to reduce this expression using the fact that $20\choose k$ = $20\choose20-k$.
Going by the expansion of $(1/4+3/4)^{20}$, we get
$$(1/4+3/4)^{20}=1= \sum_{k=0}^{20}{{20\choose k}(1/4)^k (3/4)^{20-k}}$$ $$1= \Bigr(\sum_{k=0}^{9}{{20\choose k}\bigr((1/4)^k (3/4)^{20-k}\bigl)+\bigr((1/4)^{20-k} (3/4)^{k}\bigl)}\Bigr)+ {20\choose 10}(1/4)^{10} (3/4)^{10}$$
I am stuck over here. How to simplify this from here?
Any hint would be a great help!

Answer 1

I would think that an obvious point would be to combine $\left(\frac{1}{4}\right)^k\left(\frac{3}{4}\right)^{20-k}$ as $\frac{3^{20- k}}{4^{20}}$.

Answer 2

Consider that the terms of the binomial expansion $$ \Pr (m\;\left| {\;n} \right.,p) = \left( \matrix{ n \cr m \cr} \right)p^{\,m} q^{\,n - m} $$ reperesent the Binomial distribution which is the probability of having $m$ successes in $n$ Bernoulli trials, with success probability $p$ and failure probability $q= 1-p$..

Therefore the partial sum represents the corresponding Cumulative distibution function $$ Q(m\;\left| {\;n} \right.,p) = \sum\limits_{k = 0}^m {\left( \matrix{ n \cr k \cr} \right)p^{\,k} q^{\,n - k} } $$ which is known to be expressible through the Regularized Incomplete Beta function $$ \eqalign{ & Q(m\;\left| {\;n} \right.,p) = \sum\limits_{k = 0}^m {\left( \matrix{ n \cr k \cr} \right)p^{\,k} q^{\,n - k} } = I_{\,q} (n - m,m + 1) = \cr & = \left( {n - m} \right)\left( \matrix{ n \cr m \cr} \right) \int_0^q {t^{\,n - m - 1} \left( {1 - t} \right)^{\,m} dt} \cr} $$

So in your case we have $$ \sum\limits_{k = 0}^{10} {\left( \matrix{ 20 \cr k \cr} \right)p^{\,k} q^{\,20 - k} } = 10\left( \matrix{ 20 \cr 10 \cr} \right)\int_0^{3/4} {t^{\,9} \left( {1 - t} \right)^{\,10} dt} $$ and the value can be derived from computing by a CAS the incomplete Beta to obtain $$ \sum\limits_{k = 0}^{10} {\left( \matrix{ 20 \cr k \cr} \right)p^{\,k} q^{\,20 - k} } \approx \left( \matrix{ 20 \cr 10 \cr} \right) \cdot 5.39 \cdot 10^{\, - 6} \approx 0.996 $$