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The coefficient of  x50 in the binomial expansion of


(1+x)^{1000}+x(1+x)^{999}+x^{2}(1+x)^{998}+......+x^{1000}\; is\; :

 

  • Option 1)

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

  • Option 2)

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

  • Option 3)

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

  • Option 4)

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

 

Answers (1)

best_answer

As we have learned 

Expression of Binomial Theorem -

\left ( x+a \right )^{n}= ^{n}\! c_{0}x^{n}a^{0}+^{n}c_{1}x^{n-1}a^{1}+^{n}c_{2}x^{n-2}a^{2}x-----^{n}c_{n}x^{0}a^{n}

 

- wherein

for n  +ve integral .

 

 Series given in a G.P with common ratio = \frac{x}{1+x}

No. of terms = 1001 

So sum = ({1+x} )^{1000} \frac{(\frac{x}{1+x} )^{1001}-1 }{(\frac{x}{1+x} )-1 }

 

\frac{({1+x} )^{1000} - (1+x)^{1001})}{-1}

({1+x} )^{1000} - (1+x)^{1001}

Now coeffiecient of  x^{50} = ^{1001}C_{50 }= \frac{(1001)!}{(50)!(951)!}

 

 

 

 

 


Option 1)

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

Option 2)

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

Option 3)

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

Option 4)

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

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gaurav

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