1-x-xⁿ+xⁿ+1}, where n is a natural number, is divisible by

Correct Answer is : (1−x)2(1 - x)²(1−x)2

Correct Answer is : (1−x)2(1 - x)²(1−x)2

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UPSC CDS 2 2020 Math Paper

Question : 39 of 100

Marks: +1, -0

1-x-x^{n}+x^{n+1}

, where n is a natural number, is divisible by

Solution:

Concept:

Factor the expression to identify a perfect square factor.

Explanation:

Start with the expression:

1 - x - x^{n} + x^{n+1}

Group terms:

(1 - x) - x^{n}(1 - x)

Factor out

(1 - x)

(1 - x)(1 - x^{n})

Now factor

(1 - x^{n})

using the identity:

1 - x^{n} = (1 - x)(1 + x + x^{2} + \dots + x^{n-1})

, valid for any natural number

n

Thus the entire expression becomes:

(1 - x) \cdot (1 - x) \cdot (1 + x + \dots + x^{n-1}) = (1 - x)^{2}(1 + x + \dots + x^{n-1})

Therefore, it is always divisible by

(1 - x)^{2}

, independent of the value of

n

Answer:

Option B:

(1 - x)^{2}

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