Let pn-xxn-1-n- an-1-ann-1 is divisible by x-a 2 for

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Sum of n Terms

Let pn=xxn-...

Question

Let $p (n) = x (x^{n - 1} - n \cdot a^{n - 1} + a^{n} (n - 1))$ is divisible by ${(x - a)}^{2}$ for

n > 1

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n > 2

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\forall n \in N

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None of these

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Solution

The correct option is C None of these
$p (1) = x$
Hence it is not divisible by $(x - a)^{2}$ .
$p (2) = x (x - 2 a + a^{2})$
$= x^{2} - 2 a x + a^{2} x$
$= (x - a)^{2} - a^{2} + a^{2} (x)$
$= (x - a)^{2} + a^{2} (x - 1)$ ...(i)
Hence $p (2)$ is also not divisible by $(x - a)^{2}$ .
Now ${1, 2}$ falls in the set of first n natural numbers.
Hence the correct answer is Option D.

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Let p(n)=x(xn−1−n⋅an−1+an(n−1)) is divisible by (x−a)2 for

The correct option is C None of thesep(1)=xHence it is not divisible by (x−a)2.p(2)=x(x−2a+a2)=x2−2ax+a2x=(x−a)2−a2+a2(x)=(x−a)2+a2(x−1) ...(i)Hence p(2) is also not divisible by (x−a)2.Now {1,2} falls in the set of first n natural numbers.Hence the correct answer is Option D.

Let $p (n) = x (x^{n - 1} - n \cdot a^{n - 1} + a^{n} (n - 1))$ is divisible by ${(x - a)}^{2}$ for