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UPSC NDA I 2019 Mathematics Solved Paper

Question : 21 of 120

Marks: +1, -0

\{(2x - 3y)^{2} (2x + 3y)^{2}\}^{2}

Solution:

Concept:

The number of terms in the binomial expansion of

(a + b)^n

n + 1

Explanation:

First simplify the given expression:

\{(2x - 3y)^2 (2x + 3y)^2\}^2

Rewrite as

\{ (2x - 3y) \times (2x + 3y) \times (2x - 3y) \times (2x + 3y) \}^2

Use the identity

(a - b)(a + b) = a^2 - b^2

for the product of the first two factors and the last two factors.

This gives

\{ (4x^2 - 9y^2) \times (4x^2 - 9y^2) \}^2 = \{ (4x^2 - 9y^2)^2 \}^2 = (4x^2 - 9y^2)^4

Now we have

(a + b)^n

with

a = 4x^2

b = -9y^2

, and

n = 4

The number of terms is

n + 1 = 4 + 1 = 5

Answer:

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