Question

If for $n \ge 1, \,\,\, P_{n} = \int\limits^{e}_{1}\left(log \,x\right)^{n} dx,$ then $P_{10} - 90P_{8}$ is equal to :

• A. -9
• B. 10e
• C. -9e
• D. 10

Answer 1

Answer: (C)

$P_{n} = \int\limits^{e}_{1}\left(log \,x\right)^{n} . Idx$
Integrate by parts
$P_{n} = \left(x\left(log\,x\right)^{n}\right)^{e}_{1} - \int\limits^{e}_{1} x\,n\left(log\,x\right)^{n-1} . \frac{1}{x}dx$
$P_{n}= e - n \,P_{n-1} \,\Rightarrow \, p_{n} + n\, P_{n-1} = e$
put $n = 10 \,P_{10}+ 10P_{9}= e\quad\quad \ldots\left(1\right)$
$n = 9 \quad P_{5} + 9P_{8} = e\quad\quad ...\left(2\right)$
use $\left(2\right) in \left(1\right)\quad P_{10} + 10 \left(e - 9P_{8}\right) = e$
$P_{10} - 90 \,P_{8} = e - 10 \,e$
$= - 9e$