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Question

The value of p, for which coefficient of x50 in the expression (1+x)1000+2x(1+x)999+3x2(1+x)998+....+1001x1000 is equal to 1002Cp, is

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Solution

Let S=1000r=0(1+r)xr(1+x)1000r

So, Sx(1+x)S=(1000r=0xr(1+x)1000r)1001x1001(1+x)=((1+x)1000(1(x1+x)1001)1x1+x)1001x10011+x

S(1x1+x)=((1+x)1001x1001)1001x10011+x
S=(1+x)((1+x)1001x1001)1001x1001=(1+x)10021002x1001x1002

So, S=(1002r=01002Crxr)1002x1001x1002

So, coefficient of x50 in the expansion will be, 1002C50

So, p=50 or p=100250=952 (nCr=nCnr).

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