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

\((1+x)^{101}\left(1+x^{2}-x\right)^{100}\) के \(x\) की घातों में प्रसार में पदों की संख्या है

  1. A \(302\)
  2. B \(301\)
  3. C \(202\)
  4. D \(101\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(202\)

Step-by-step Solution

Detailed explanation

Given expansion is \((1+x)^{101}\left(1-x+x^{2}\right)^{100}\) \(=(1+x)(1+x)^{100}\left(1-x+x^{2}\right)^{100}\) \(=(1+x)\left[(1+x)\left(1-x+x^{2}\right)\right]^{100}\) \(=(1+x)\left[\left(1-x^{3}\right)^{100}\right]\) Expansion \(\left(1-x^{3}\right)^{100}\) will have…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app