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Mathematics Question on binomial expansion formula

The coefficient of x50x^{50} in the binomial expansion of (1+x)1000+x(1+x)999+x2(1+x)998+....+x1000(1 + x)^{1000} + x (1 + x)^{999} + x^2(1 + x)^{998} + .... + x^{1000} is:

A

(1000)!(50)!(950)!\frac{(1000)!}{(50)! ( 950)!}

B

(1000)!(49)!(951)!\frac{(1000)!}{(49)! ( 951)!}

C

(1001)!(51)!(950)!\frac{(1001)!}{(51)! ( 950)!}

D

(1001)!(50)!(951)!\frac{(1001)!}{(50)! ( 951)!}

Answer

(1001)!(50)!(951)!\frac{(1001)!}{(50)! ( 951)!}

Explanation

Solution

Let given expansion be S=(1+x)1000+x(1+x)999+x2(1+x)998+...+...+x1000S = \left(1+x\right)^{1000} + x \left(1+x\right)^{999} + x^{2} \left(1+x\right)^{998} + ... + ...+ x^{1000} Put 1 + x = t S=t1000+xt999+x2(t)998+...+x1000 S= t^{1000} + xt^{999} + x^{2} \left(t\right)^{998} + ...+x^{1000} This is a G.P with common ratio xt\frac{x}{t} S=t1000[1−(xt)1001]1−xtS = \frac{t^{1000} \left[ 1 - \left(\frac{x}{t}\right)^{1001}\right]}{1- \frac{x}{t}} =(1+x)1000[1−(x1+x)1001]1−x1+x=\frac{\left(1+x\right)^{1000} \left[1- \left(\frac{x}{1+x}\right)^{1001}\right]}{ 1 - \frac{x}{1+x}} =(1+x)1001[(1+x)1001−x1001](1+x)1001= \frac{\left(1+x\right)^{1001} \left[\left(1+x\right)^{1001} -x^{1001}\right]}{\left(1+x\right)^{1001}} =[(1+x)1001−x1001]= \left[\left(1+x\right)^{1001} - x^{1001}\right] Now coeff of x50x^{50} in above expansion is equal to coeff of x50x^{50} in (1+x)1001(1 + x)^{1001} which is 1001C50{^{1001}C_{50}} =(1001)!50!(951)!= \frac{\left(1001\right)!}{50! \left(951\right)!}