For all positive integers n 1, x(xn-1-nan-1) + an(n-1) is divisible by

Question

For all positive integers n > 1, x(xn-1-nan-1) + an(n-1) is divisible by (A) (x - a)2 (B)  x - a  (C)  2 ( x - a) (D) x + a

Tardigrade Admin · Accepted Answer

Let P(n): x(xn-1 - nan-1) + an(n-1), where n> 1 P(2): x(x-2a) + a2 = (x- a)2 P(3): x (x2 -3a2) + 2a3 = x3 -3a2x +2a3 ⇒ P(3) =(x-a)(x2 + ax-2a2) which is divisible by (x - a) Hence P(n) is divisible by (x - a).

Q. For all positive integers $n > 1$ ,
${x (x^{n - 1} - n a^{n - 1}) + a^{n} (n - 1)}$ is divisible by

$(x - a)^{2}$

$x - a$

$2 (x - a)$

$x + a$

Q. For all positive integers n>1, {x(xn−1−nan−1)+an(n−1)} is divisible by

(x−a)2

x−a

2(x−a)

x+a

Let P(n):x(xn−1−nan−1)+an(n−1), where n>1 P(2):x(x−2a)+a2=(x−a)2 P(3):x(x2−3a2)+2a3=x3−3a2x+2a3 ⇒P(3)=(x−a)(x2+ax−2a2) which is divisible by (x−a) Hence P(n) is divisible by (x−a).

Q. For all positive integers $n > 1$ ,
${x (x^{n - 1} - n a^{n - 1}) + a^{n} (n - 1)}$ is divisible by

$(x - a)^{2}$

$x - a$

$2 (x - a)$

$x + a$