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Geometric Progression

Prove that x ...

Question

Prove that x $(x^{n - 1} - n a^{n - 1}) + a^{n} (n - 1)$ is divisible by $(x - a)^{2}$ for all positive integers n greater than 1

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Solution

Let $A_{n} = x (x^{n - 1} - n a^{n - 1}) + a^{n} (n - 1)$
For n = 2, $A_{2} = x (x - 2 a) + a^{2}$
$= x^{2} - 2 a x a^{2} = (x - a)^{2} .$
Thus $A_{2}$ is divisible by $(x - a)^{2} .$
Now assume that $A_{m}$ is divisible by $(x - a)^{2}$ for m $\geq$ 2, that is, assume
$A_{m} = x (x^{m - 1} - m a^{m - 1}) + a^{m} (m - 1)$
$= (x - a)^{2}$ f (x)
so that $x^{m} = m x a^{m - 1} - a^{m} (m - 1) + (x - a)^{2} f (x)$
We then have
$A_{m - 1} = x [x^{m} - (m + 1) a^{m}] + a^{m + 1} m$
$= x . x^{m} - (m + 1) x a^{m} + m a^{m, + 1}$
$= x [m x a^{m - 1} - a^{m} (m - 1) + (x - a)^{2} f (x)]$
$- (m + 1) x a^{m} + m a^{m + 1}$ by
$= m a^{m - 1} [(x^{2} - 2 x a + a^{2}) + x (x - a)^{2} f (x)]$
$= m a^{m - 1} ((x - a)^{2} + x (x - a)^{2} f (x)$
$= (x - a)^{2} [m a^{m - 1} + x f (x)],$
After simplification.
This shows that $A_{m + 1}$ is divisible by $(x - a)^{2} .$
Hence by induction, $A_{n}$ is divisible by $(x - a)^{2}$ for all positive integers n > 1.

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