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OpenStudy (anonymous):

help me in this problem

OpenStudy (anonymous):

Find out in terms of n \[\sum_{r=1}^{n}r ^{2}* nCr\]

OpenStudy (anonymous):

we have to find a closed form right ?

OpenStudy (anonymous):

yeah

OpenStudy (anonymous):

hm $$\large 2^{n-2} \times n \times (n+1)$$

OpenStudy (anonymous):

how??

OpenStudy (anonymous):

You can verify by induction .. I guessed it by chekcing up few terms .. ;)

OpenStudy (anonymous):

i guess the answer is somewhat different..

OpenStudy (anonymous):

what is the answer ?

OpenStudy (anonymous):

wait a while

OpenStudy (anonymous):

hmmm.. right u r but i wanna know the exact process..

OpenStudy (anonymous):

ok, i solved it myself right now... thanx a lot, @ foolformath :)

OpenStudy (anonymous):

You are welcome, how you solve it ?

OpenStudy (anonymous):

n^2=n(n-1)+n

OpenStudy (anonymous):

what is that ?

OpenStudy (anonymous):

it will help u to get the expression. anyway, bye for now.. will post the solution later..

OpenStudy (anonymous):

duh ...

OpenStudy (anonymous):

howz that help you ?

OpenStudy (anonymous):

you derived the closed form based on that ?

OpenStudy (anonymous):

yeah

OpenStudy (anonymous):

u can also do it by differenciating

OpenStudy (anonymous):

but dont worry, i will be back and will post the solution..

OpenStudy (anonymous):

Sure, When you do just post the one you did with using $$ n^2=n(n-1)+n$$ and the derivative one is pretty obvious, expand (1+x)^n then do derivative with respect to x after that multiply with x and then again do the derivative you will left with with something like this $$ (-1+n) n x (1+x)^{-2+n}+n (1+x)^{-1+n}$$ in the LHS and subsitute x =1 you will have your required closed form in both RHS and LHS.

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