A) \[\frac{\left( 1000 \right)!}{\left( 50 \right)!\left( 950 \right)!}\]
B) \[\frac{\left( 1000 \right)!}{\left( 49 \right)!\left( 951 \right)!}\]
C) \[\frac{\left( 1001 \right)!}{\left( 51 \right)!\left( 950 \right)!}\]
D) \[\frac{\left( 1001 \right)!}{\left( 50 \right)!\left( 951 \right)!}\]
Correct Answer: D
Solution :
Let given expansion be \[S={{(1+x)}^{1000}}+x{{(1+x)}^{999}}+{{x}^{2}}{{(1+x)}^{998}}+...\] \[+...+{{x}^{1000}}\] Put 1 + x = t \[S={{t}^{1000}}+x{{t}^{999}}+{{x}^{2}}{{(t)}^{998}}+...+{{x}^{1000}}\] This is a G.P with common ratio\[\frac{x}{t}\] \[S=\frac{{{t}^{1000}}\left[ 1-{{\left( \frac{x}{t} \right)}^{1001}} \right]}{1-\frac{x}{t}}\] \[\frac{{{(1+x)}^{1000}}\left[ 1-{{\left( \frac{x}{1+x} \right)}^{1001}} \right]}{1-\frac{x}{1+x}}\] \[=\frac{{{(1+x)}^{1000}}\left[ {{(1+x)}^{1001}}-{{x}^{1001}} \right]}{{{(1+x)}^{1001}}}\] \[=[{{(1+x)}^{1001}}-{{x}^{1001}}]\] Now coeff of x50 in above expansion is equal to coeff of \[{{x}^{50}}\]in \[{{(1+x)}^{1001}}\]which is \[^{1001}{{C}_{50}}\] \[=\frac{(1001)!}{50!(951)!}\]
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