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UPSC NDA 2 2019 Mathematics Solved Paper

Question : 24 of 120

Marks: +1, -0

(1 + 2x + x^{2})^{5} + (1 + 4y + 4y^{2})^{5}

Solution:

Concept:

A binomial expansion

(a+b)^n

has

n+1

terms.

Explanation:

First, simplify each bracket.

1 + 2x + x^2 = (1+x)^2

, so

(1+2x+x^2)^5 = ((1+x)^2)^5 = (1+x)^{10}

Similarly,

1 + 4y + 4y^2 = (1+2y)^2

, so

(1+4y+4y^2)^5 = ((1+2y)^2)^5 = (1+2y)^{10}

Thus, the given expression becomes

(1+x)^{10} + (1+2y)^{10}

Each binomial expansion has

10+1 = 11

terms.

The constant term

1

appears in both expansions, so it is counted twice.

Therefore, total distinct terms

= 11 + 11 - 1 = 21

Answer:

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