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$\sum_{r=0}^n {nCr} (-1/2)^{3r} {4}^{n+r} = -240$ 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

Hydrous Caperilla
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  • You might need to typeset this question better. It is hard to read. – J. Redman Jun 25 '18 at 01:35
  • Should the summation start with $r=0$? But if so, then the left-hand side is $2^n$ which is not negative unless you are allowing for $n$ to be a complex number. Is there something else wrong with the equation? – JimB Jun 25 '18 at 02:02
  • The equation is starting with $r=1$ in my question and the answer given for n is 4 – Hydrous Caperilla Jun 25 '18 at 02:15
  • You should edit your question to reflect the fact that the summation variable is $r$, not $i$ and the lower limit is $1$. Just below the question is a button edit for this purpose. – Ross Millikan Jun 25 '18 at 02:56

2 Answers2

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$$\sum_{r=0}^n \binom {n}{r}(-1)^r2^{2n-r}=2^{2n}\sum_{r=0}^n \binom {n}{r} (-1)^r2^{-r}=$$ $$=2^{2n}(-1)^r(2^{-1})^r=$$ $$=2^{2n}\sum_{r=0}^n \binom {n}{r}((-1)2^{-1})^r=$$ $$=2^{2n}(1+((-1)2^{-1})^n=$$ $$=2^{2n}\left(1-\frac {1}{2}\right)^n =2^{2n}\left(\frac {1}{2}\right)^n=$$ $$=2^{2n}2^{-n}=2^n \neq -240.$$

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. $

DanielWainfleet
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$240$ is such a small number in terms of combinations that I made a spreadsheet and tried by hand. I find that if the sum starts from $r=1$ and $n=4$ the equation is satisfied. As you didn't show your work I can't tell if you went wrong in assuming $r$ starts from $0$ or somewhere else.

Ross Millikan
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