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

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

Principle of Mathematical Induction

Solution:

Let $P\left(n\right): x\left(x^{n-1} - na^{n-1}\right) + a^{n}\left(n-1\right)$, where $n> 1$
$P\left(2\right) : x\left(x-2a\right) + a^{2} = \left(x- a\right)^{2}$
$ P\left(3\right): x \left(x^{2} -3a^{2}\right) + 2a^{3} = x^{3} -3a^{2}x +2a^{3}$
$\Rightarrow P\left(3\right) =\left(x-a\right)\left(x^{2} + ax-2a^{2}\right) $
which is divisible by $\left(x - a\right)$
Hence $P\left(n\right)$ is divisible by $\left(x - a\right)$.