Is there any Simplification to the sum 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!
 A: 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
$$
\sum\limits_{k = 0}^{10} {\left( \matrix{  20 \cr   k \cr}  \right)p^{\,k} q^{\,20 - k} }
  \approx \left( \matrix{  20 \cr  10 \cr}  \right) \cdot 5.39 \cdot 10^{\, - 6}  \approx 0.996
$$
A: 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 $\frac{3^{20- k}}{4^{20}}$.
