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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 Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
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