Binomial Q..
\[\left(\begin{matrix}1000 \\ 50\end{matrix}\right) +\left(\begin{matrix}999 \\ 49\end{matrix}\right)+\left(\begin{matrix}998 \\ 48\end{matrix}\right)+.......+\left(\begin{matrix}949 \\ 0\end{matrix}\right)\]=?
@ganeshie8
@ParthKohli
Hockey stick, eh? :P
no.. this is what i got as the exp. while solving a problem..
and thats not the answer..
There seems to be an issue here 1000 to 949 are 52 numbers 50 to 0 are 51 numbers
Oh haha, good observation.
oops! thanks! its 950
Hockey Stick Identity http://www.artofproblemsolving.com/wiki/index.php/Combinatorial_identity
pls tell me how to find the sum..
The answer is then simply \(\dbinom{1001}{50}\)
but how?
First look at what the identity says we can bother about proving afterwards
you must be familiar with pascal's triangle and how it relates to the binomial coefficients, right ?
yes..
Good. Add the blue numbers, what do you get ? |dw:1456992132781:dw|
oh ok.. i am reading from the link...pls wait..
take your time
its a bit confusing.. can you write the general formula and the one for the above exp. to make things clearer..
Forget about the link. The identity is easy if you interpret it using the pascal's triangle
yes.. i understood the actual meaning of the identity...that is if we add all the no's in a diagonal in the pascal's triangle..the sum will be equal to the number directly perpendicular to it.. and this shape looks like a hockey stick. right?
But i didn't understand the mathematical formulation of it..
\[\sum\limits_{i=r}^n \dbinom{i}{i-r} = \dbinom{n+1}{n-r}\] plugin \(n=1000,~~r = 950\)
why i=r wouldn't the "i-r" term be zero always?
@ParthKohli ?
Here \(i\) is the only variable \(n\) and \(r\) are fixed
ok now i got it! thanks @ganeshie8 !
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