Question

Find the value of
${}^1{P}_1+2{\cdot }^2{P}_2+3{\cdot }^3{P}_3+4{\cdot }^4{P}_4+...+n^n{P}_n$

• A. ${}^{n+1}{P}_{n+1}$
• B. ${}^{n+1}{P}_{n+1}-1$
• C. ${}^{n+1}{P}_{n+1}-2$
• D. ${}^{n+1}{P}_{n+1}+1$

Answer 1

The correct option is B ${}^{n+1}{P}_{n+1}-1$

$r^{th}$ term of the given series
$r{\cdot }^r{P}_r=r\cdot r!=\left(\right(r+1)-1)!r!$
$=(r+1)!-r!$
On putting $r=1,2,3,....n$ and adding, we get
${}^1{P}_1+2{\cdot }^2{P}_2+3{\cdot }^3{P}_3+...+n{\cdot }^n{P}_n$
$=(2!-1!)+(3!-2!)+(4!-3!)+...+(n+1)!-n!$
$=(n+1)-1!$
${=}^{n+1}{P}_{n+1}-1$