Question
Let $$S = \sum\limits_{k = 0}^{n - 1} {^{k + 2}{P_2},}\, $$ then
A.
$$n$$ divides $$3S$$
B.
$$n + 1$$ divides $$3S$$
C.
$$n + 2$$ divides $$3S$$
D.
All are correct
Answer :
All are correct
Solution :
We have, $$S = \sum\limits_{k = 0}^{n - 1} {^{k + 2}{P_2}} = 2!\sum\limits_{k = 0}^{n - 1} {^{k + 2}{C_2}} $$
$$\eqalign{
& = \left( 2 \right)\left[ {^2{C_2} + {\,^3}{C_2} + {\,^4}{C_2} + ..... + {\,^{n + 1}}{C_2}} \right] \cr
& {\text{But,}}{{\text{ }}^2}{C_2} + {\,^3}{C_2} + {\,^4}{C_2} + ..... + {\,^{n + 1}}{C_2} \cr
& = \left( {^3{C_3} + {\,^3}{C_2}} \right) + {\,^4}{C_2} + ..... + {\,^{n + 1}}{C_2} \cr
& = \left( {^4{C_3} + {\,^4}{C_2}} \right) + {\,^5}{C_2} + ..... + {\,^{n + 1}}{C_2} \cr
& = {\,^5}{C_3} + {\,^5}{C_2} + ..... + {\,^{n + 1}}{C_2} \cr
& = {\,^{n + 2}}{C_3} = \frac{1}{6}n\left( {n + 1} \right)\left( {n + 2} \right) \cr
& \Rightarrow 3S = n\left( {n + 1} \right)\left( {n + 2} \right) \cr} $$
$$ \Rightarrow n$$ divides $$3S,\left( {n + 1} \right)$$ divides $$3S$$ and $$\left( {n + 2} \right)$$ divides $$3S.$$