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UPSC NDA I and II 2020 Math Solved Paper

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Question : 31 of 120

Marks: +1, -0

If

(1 + 2x - x^{2})^{6} = a_{0} + a_{1}x + a_{2}x^{2} + ... + a_{12}x^{12}

a_{0} - a_{1} + a_{2} - a_{3} + a_{4} - ... + a_{12}

equal to?

32
64
2048
4096

Solution:

Concept:

Substituting

x = -1

converts the alternating sum into a direct evaluation of the polynomial.

Explanation:

We are given

(1 + 2x - x^{2})^{6} = a_{0} + a_{1}x + a_{2}x^{2} + \dots + a_{12}x^{12}

.

To find

a_{0} - a_{1} + a_{2} - a_{3} + \dots + a_{12}

, set

x = -1

.

Then each term

a_{k}x^{k}

becomes

a_{k}(-1)^{k}

, yielding the desired alternating sum.

Compute:

(1 + 2(-1) - (-1)^{2})^{6} = (1 - 2 - 1)^{6} = (-2)^{6} = 64

.

Thus,

a_{0} - a_{1} + a_{2} - a_{3} + \dots + a_{12} = 64

.

Answer:

64

(Option B).

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