NIMCET 2023 Question Paper with solutions

Question : 37 of 120

Marks: +1, -0

The coefficient of

x^{50}

in the expansion

(1+x)^{1000}+2 x(1+x)^{999}+3 x^2(1+x)^{998}+. . .

+1001 x^{1000}

is:

Solution:

Let

{\S}=(1+x)^{1000}+2 x(1+x)^{999}+3 x^2 (1+x)^{998}+. . .+1000 x^{999}(1+x)+1001 x^{1000}

Above is a G.P. with common ratio,

r=\;{x}/{1+x}

∴ [\;{x}/{(1+x)}] {\S}=x(1+x)^{999}+2 x^2(1+x)^{998} +. . .+1000 ⋅ x^{1000}+\;{1001 x^{1001}}/{1+x}

Subtracting,

(1-\;{x}/{1+x}) {\S}

=(1+x)^{1000}+x(1+x)^{999}+x^2(1+x)^{998} +. . .+x^{1000}-\;{1001 x^{1001}}/{1+x}

\;\text " Or, "\; {\S}=(1+x)^{1001}+x(1+x)^{1000}+x^2(1+x)^{999} +. . .+x^{1000}(1+x)-1001 x^{1001}

=\;{(1+x)^{1001} [1-(x-(1+x))^{1001}] }/{1-x}-1001 x^{1001}

Sum of the G.P.

\;=(1+x)^{1002} [1- (\;{x}/{(1+x)}) ^{1001}] -1001 x^{1001}

\;=(1+x)^{1002}-x^{1001}(1+x)-1001 x^{1001}

\;=(1+x)^{1002}-x^{1002}-1002 x^{1001}

Now the coefficients of

x^{50}

on the R.H.S. of (i)

={ }^{1002} {\C}_{50} \;\text ". "\;

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