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Coeff. of $x^{50}$ in $(1+x)^{1000}$ + $x(1+x)^{999}$ + $x^2(1+x)^{998}$ + .... + $x^{1000}$ is

$\begin{array}{1 1} ^{1000} C_50 \\ ^{1001}C_{50} \\ ^{1001}C_{51} \\ ^{1000}C_{51} \end{array} $

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$(1+x)^{1000}+x(1+x)^{999}+x^2(1+x)^{998}+......x^{1000}$ is a G.P.
$\therefore $ Sum of $1000 $ terms of this G.P. is
$\therefore\:$ Coeff. of $x^{50} $ in the given G.P. is $^{1001}C_{50}$
answered Nov 23, 2013 by rvidyagovindarajan_1

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