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

\((1-x)^{101}\left(x^{2}+x+1\right)^{100}\) નાં વિસ્તરણમાં \(x^{256}\) નો સહગુણક મેળવો.

  1. A \({-}^{100} \mathrm{C}_{16}\)
  2. B \(^{100} \mathrm{C}_{16}\)
  3. C \(^{100} \mathrm{C}_{15}\)
  4. D \(-{ }^{100} \mathrm{C}_{15}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(^{100} \mathrm{C}_{15}\)

Step-by-step Solution

Detailed explanation

\(y=(1-x)(1-x)^{100}\left(x^{2}+x+1\right)^{100}\) \(y=(1-x)\left(x^{3}-1\right)^{100}\) \(y=\left(x^{3}-1\right)^{100}-x\left(x^{3}-1\right)^{100}\) Coff. Of \(x^{256}\) in \(y=-\) coff of \(x^{255}\) in \(\left(x^{3}-1\right)^{100}\)…
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