# The sum of the series

Question:

The sum of the series $\sum_{r=0}^{10}{ }^{20} \mathrm{C}_{f}$ is ____________

Solution:

for $\sum_{r=0}^{10}{ }^{20} C_{r}$

for $n=2$

consider,

$(1+x)^{20}={ }^{20} C_{0}+{ }^{20} C_{1} x+\ldots \ldots+{ }^{20} C_{20} x^{20}$

i. e. $\sum_{r=0}^{10}{ }^{20} C_{r}={ }^{20} C_{0}+{ }^{20} C_{1}+{ }^{20} C_{2}+\ldots \ldots+{ }^{20} C_{10}$

Since ${ }^{20} C_{1}={ }^{20} C_{19}$

${ }^{20} C_{2}={ }^{20} C_{18}$

${ }^{20} C_{11}={ }^{20} C_{9}$

i. e. $(1+x)^{20}=2\left({ }^{20} C_{0}+{ }^{20} C_{1} x+\ldots \ldots+{ }^{20} C_{9} x^{9}+{ }^{20} C_{10} x^{10}\right)-{ }^{20} C_{10} x^{10}$

$(1+x)^{20}=2\left(\sum_{r=0}^{10}{ }^{20} C_{r} x^{r}\right)-{ }^{20} C_{10} x^{10}$

$2\left(\sum_{r=0}^{10}{ }^{20} C_{r} x^{r}\right)$

put $x=1$

i.e. $2 \sum_{r=0}^{10}{ }^{20} C_{r}=(1+1)^{20}+{ }^{20} C_{10}$

i.e. $\sum_{r=0}^{10}{ }^{20} C_{r}=2^{19}+\frac{{ }^{20} C_{10}}{2}$