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# find the coefficient of x^98 in (x-1)(x-2)(x-3)...(x-99)(x-100)?

10 years ago

If we expand (x-1)(x-2).......(x-100) we will get
(x-1)(x-2).......(x-100)=x100 -(1+2+3....100)x99+(Sum of product of all first 100 integers taken two at a time)x98+................

Now we have to find "Sum of product of all first 100 integers taken two at a time" which is the required co-efficient.

(1+2+3+....100)2=12+22+32+.....+1002+2(Sum of product of all first 100 integers taken two at a time)

So  Sum of product of all first 100 integers taken two at a time=((1+2+3+....100)2-(12+22+32+.....+1002))/2

Now use 1+2+3+....+n=n(n+1)/2 and 12+22+32+.....+n2=n(n+1)(2n+1)/6

Here n=100

So we get the required co-efficient=12582075
all the best
--
regards
Ramesh
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