Question

If $f\left(n\right)=1\times 2\times 3\times \times (n-1)\times n$, where n is a natural number, then $\sum _{r=1}^{20}rf\left(r\right)=f\left(x\right)-1$, then the value of $x$ is?

• A. $25$
• B. $21$
• C. $28$
• D. $24$

Answer 1

The correct option is B $21$

here function

$f\left(n\right)=1\times 2\times 3\times .........\times (n-1)]\times n$

$f\left(n\right)=n!$ $...................................\left(1\right)$

$so,n.f\left(n\right)=n\times 1\times 2\times 3\times .........\times (n-1)]\times n$

$n.f\left(n\right)=n.n!$

$=(n+1-1).n!$

$n.f\left(n\right)=(n+1)!-n!$

$here{\sum }_{r=1}^{20}rf\left(r\right)={\sum }_{r=1}^{20}\left[\right(r+1)!-r!]$

$=(2!-1!)+(3!-2!)+(4!-3!)+......+(20!-19!)+(21!-20!)$

$=21!-1!\left(simplificationform\right)$

$=I\left(21\right)-f\left(1\right)\left[by\right(1\left)\right]$

$=f\left(21\right)-1\left(f\right(1)=1)$

but ${\sum }_{r=1}^{20}rf\left(r\right)=f\left(x\right)-1$

So $x=21$