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JEE Mains · Maths · STD 11 - 7. binomial theoram

\((1+x)^{1000}+x(1+x)^{999}+x^{2}(1+x)^{998}+\) \(\cdots \cdots+x^{1000}\) के द्विपद प्रसार में \(x^{50}\) का गुणाँक है

  1. A \(\frac{{\left( {1000} \right)!}}{{\left( {50} \right)!\left( {950} \right)!}}\)
  2. B \(\frac{{\left( {1000} \right)!}}{{\left( {49} \right)!\left( {951} \right)!}}\)
  3. C \(\frac{{\left( {1001} \right)!}}{{\left( {51} \right)!\left( {950} \right)!}}\)
  4. D \(\frac{{\left( {1001} \right)!}}{{\left( {50} \right)!\left( {951} \right)!}}\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

Let given expansion be \(\mathrm{S}=(1+x)^{1000}+x(1+x)^{999}+x^{2}\) \((1+x)^{998}+\ldots+\ldots+x^{1006}\) Put \(1+x=t\) \(\mathrm{S}=t^{1000}+x t^{999}+x^{2}(t)^{998}+\ldots+x^{1000}\) This is a \(G.P\) with common ratio \(\frac{x}{t}\)…
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