$\sum_{r=0}^n {nCr} (-1/2)^{3r} {4}^{n+r} = -240$ then $n$ is equal to
My attempt
I rearranged the formulae to $\sum_{r=0}^n {n\choose r} (-1)^r {2}^{2n-r} $ and then converted this into binomial expansion but I am getting the whole expansion to be zero which is not possible so where am I going wrong? I took $2^n$ outside and rest becomes a binomial coefficient but the whole summation becomes zero in this process
$$\sum_{r=0}^n \binom {n}{r}(-1)^r2^{2n-r}=2^{2n}\sum_{r=0}^n \binom {n}{r} (-1)^r2^{-r}=$$ $$=2^{2n}(-1)^r(2^{-1})^r=$$ $$=2^{2n}\sum_{r=0}^n \binom {n}{r}((-1)2^{-1})^r=$$ $$=2^{2n}(1+((-1)2^{-1})^n=$$ $$=2^{2n}\left(1-\frac {1}{2}\right)^n =2^{2n}\left(\frac {1}{2}\right)^n=$$ $$=2^{2n}2^{-n}=2^n \neq -240.$$
On the other hand if the summation is from $r=1$ to $r=n$ instead of from $r=0$ to $r=n$ then the sum is $$2^n- \binom {n}{0}(-1)^02^{2n-0}=2^n-4^n.$$ And $2^n-4^n=-240\iff$ $ (2^n)^2-2^n+240=0\iff$ $ (2^n -16 )(2^n+15)=0\iff$ $\iff 2^n-16=0$ (...because $2^n+15>0$....) iff $n=4. $