Question

The value of $(2\cdot {}^1{P}_0-3\cdot {}^2{P}_1+4\cdot {}^3{P}_2-\cdots \cdots \text{up to}{51}^{\text{th}}\text{term})+(1!-2!+3!-4!+\cdots \cdots \text{up to}{51}^{\text{th}}\text{term})$ is equal to

• A. $1-51\left(51\right)!$
• B. $1+\left(52\right)!$
• C. $1$
• D. $1+\left(51\right)!$

Answer 1

The correct option is B $1+\left(52\right)!$

Let
$S=(2.{}^1{P}_0-3.{}^2{P}_1+4.{}^3{P}_2-\cdots \cdots \text{up to}{51}^{th}\text{term})+(1!-2!+3!-4!+\cdots \cdots \text{up to}{51}^{th}\text{term})$
$(\because {}^n{P}_{n-1}=n!)$

$\Rightarrow S=(2\times 1!-3\times 2!+4\times 3-.....+52\times 51!)+(1!-2!+3!-4!+\cdots \cdots +51!)$

$\Rightarrow S=(2!-3!+4!-.....+52!)+(1!-2!+3!-4!+\cdots \cdots +51!)$

$\therefore S=1+52!$