Question

If y = x${}^n$ log x then
$y_n=n![logx+1+\frac{1}{2}+\frac{1}{3}....+\frac{1}{n}]$
$\forall nϵN$ and x > 0.

Answer 1

For p (1) we have to prove
$y_1$ = 1! (log x + 1)
where y = x log x for n = 1
$y_1=x.\frac{1}{x}+1.$ log x = (log x + 1) = 1! (log x + 1)
Thus p(1) holds good.
Now assume $y_n=n!(logx+1+\frac{1}{2}+...+\frac{1}{n})$
where y = $x^n$ log x for n = n.
p(n + 1) = $y_{n+1}of{x}^{n+1}$ (log x)
$=y_{n}of\frac{d}{dx}\left\{x^{n+1}logx\right\}$
$=y_n\left\{\right(n+1)x^{n}logx+x^{n+1}\frac{1}{x}\}$
$=y_n\left\{\right(n+1\left)x^{n}logx\right\}+y_n\left(x^n\right)$
= (n + 1) p (n) + n!
= (n + 1) n! $[logx+1+\frac{1}{2}+...+\frac{1}{n}]n!\frac{(n+1)}{(n+1)}$
= (n + 1)! $[logx+1+\frac{1}{2}+\frac{1}{3}+...\frac{1}{n+1}]$
Hence p(n + 1) also holds good.