no. of subsets which have at least $r+1$ element is A Set has $2r+1$ elements. Then the no. of subsets which have at least $r+1$ element is
My Try::
selecting $r+1$ element from $2r+1$ which is $\displaystyle  = \binom{2r+1}{r+1}$
selecting $r+2$ element from $2r+1$ which is $\displaystyle  = \binom{2r+1}{r+2}$
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selecting $2r+1$ element from $2r+1$ which is $\displaystyle  = \binom{2r+1}{2r+1}$
So Total $ = \binom{2r+1}{r+1}+\binom{2r+1}{r+2}+..........+\binom{2r+1}{2r+1}$
and answer given is $ = 2^{2r}$
How can i Calculate it. Thanks

Substitute $x=1$ into $(1+x)^{2r+1}=\sum_{k=0}^{2r+1}\binom{2r+1}{k}x^k$ to get
$2^{2r+1}$
$=\sum_{k=0}^{2r+1}\binom{2r+1}{k}$
$=\sum_{k=0}^{r}\binom{2r+1}{k}+\sum_{k=r+1}^{2r+1}\binom{2r+1}{k}$
$=\sum_{k=0}^{r}\binom{2r+1}{2r+1-k}+\sum_{k=r+1}^{2r+1}\binom{2r+1}{k}$
$=2\sum_{k=r+1}^{2r+1}\binom{2r+1}{k}$.
