Saved Bookmarks
| 1. |
Prove that ((n^(2))!)/((n!)^(n)) is a natural number for all n in N. |
|
Answer» Solution :We have to PROVE that `(n^(2))!` is divisible by `(n!)^(n)`. We know that product of r CONSECUTIVE integers is divisible by r! Now `(n^(2))!=1xx2xx3xx4xx..XX n^(2)` `=(1xx2xx3xx..xx n)xx` `[(n+1)(n+2)xx.. .. Xx(2n)]xx` `[(2n+1)(2n+2).. .. (3N)]xx` .. .. .. .. `[(n^(2)-(n-1)(n^(2)-n).. .. n^(2)]` THUS `(n^(2))!` consists n groups of product of n consecutive integers. Each group is divisible by n!. So, `(n^(2))!` is divisible by `(n!)^(n)`. |
|