The sum of 20 terms of the series whose rth term s given by k`T(n)=(-1)^n(n^2+n+1)/(n !)` is `(20)/(19 !)` b. `(21)/(20 !)-1` c. `(21)/(20 !)` d. none of these

Options

A. `20/(19!)`
B. `(21)/(20 !)-1`
C. `(21)/(20!)`
D. none of these

Answer

B

Explanation

`T_(r)=(-1)^(r )(r^(2)+r+1)/(r!)`
`=(-1)^(r )[r/((r-1)!)+1/((r-1)!)+1/(r!)]`
`=(-1)^(r )[1/((r-2)!)+1/((r-1)!)+1/((r-1)!)+1/(r!)]`
`=[((-1)^(r ))/(r!)+((-1)^(r ))/((r-1)!)]+[((-1)^(r ))/((r-1)!)+((-1)^(r ))/((r-2)!)]`
`=[(-1)^(r )/(r!)-((-1)^(r-1))/((r-1)!)]-[((-1)^(r-1))/((r-1)!)-((-1)^(r-2))/((r-2)!)]`
=V(r )-V(r-1)
`thereforesum_(r=1)^(n)T_(r )=V(n)-V(0)=[((-1)^(n))/(n!)-((-1)^(n-1))/((n-1)!)]`-1
Therefore. the sum of 20 terms is
`[1/(20!)-(-1)/(19!)]-1=21/(20!)-1`