Question

If for $n\ge 1,P_n=\underset{1}{\overset{e}{\int }}{\left(logx\right)}^{n}dx,$ then $P_{10}-90P_8$ is equal to :

• A. -9
• B. 10e
• C. -9e
• D. 10

Answer 1

Answer: (C)

$P_n=\underset{1}{\overset{e}{\int }}{\left(logx\right)}^n.Idx$
Integrate by parts
$P_n={\left(x{\left(logx\right)}^n\right)}_1^e-\underset{1}{\overset{e}{\int }}xn{\left(logx\right)}^{n-1}.\frac{1}{x}dx$
$P_n=e-n{P}_{n-1}\Rightarrow p_n+n{P}_{n-1}=e$
put $n=10P_{10}+10P_9=e\dots \left(1\right)$
$n=9P_5+9P_8=e...\left(2\right)$
use $\left(2\right)in\left(1\right)P_{10}+10\left(e-9P_8\right)=e$
$P_{10}-90P_8=e-10e$
$=-9e$