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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Binomial Theorem

NDA 2019 · Mathematics · Binomial Theorem

How many terms are there in the expansion of (1+2 x+x^2)^5+(1+4 y+4...

How many terms are there in the expansion of (1+2x+x2)5+(1+4y+4y2)5\left(1+2 x+x^2\right)^5+\left(1+4 y+4 y^2\right)^5 ?

Options

  1. A12
  2. B20
  3. C21
  4. D22

Correct answer

C. 21

Step-by-step solution

(1+2x+x2)5+(1+4y+4y2)5=[(1+x)2]5+[(1+2y)2]5=(1+x)10+(1+xy)10∴ Total numbers of  terms in each of them is 11 but constant term will be added ∴ Total terms =11+11−1=21\begin{aligned} & \left(1+2 x+x^2\right)^5+\left(1+4 y+4 y^2\right)^5 \\ & =\left[(1+x)^2\right]^5+\left[(1+2 y)^2\right]^5 \\ & =(1+x)^{10}+(1+x y)^{10} \\ & \therefore \text{ Total numbers\text{ of } terms in each of them is } 11 \text{ but constant } \\ & \text{term will be added } \\ & \therefore \quad \text{ Total terms }=11+11-1 \\ & =21\end{aligned}
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