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Suppose n is a natural number, P rove that is always an integer.

 


Suppose n is a natural number,




Prove that    is always an integer.

Grade:10

1 Answers

Jit Mitra
25 Points
10 years ago

Suppose there are n consecutive natural numbers starting with m.

 

product = m(m+1)(m+2).....(m+n-1) = (m+n-1)!/(m-1)! = n!* m+n-1Cn

We know m+n-1Cn is an integer. Therefore any n consecutive integers are divisible by n!

Now we take (n!)!. This means product of all numbers from 1 to n!

We divide this product into rows containing n consecutive integers.

 

1            2         3            4      ............       n

n+1      n+2      n+3        n+4  ..............      2n

2n+1    2n+2    2n+3       2n+4   ...........      3n

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.

.

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n!-n+1    n!-n+2       n!-n+3       .............     n!

 

There are (n-1)! rows each containing n numbers [ n(n-1)! = n! ]

Therefore, product of numbers in each row is divisible by n!

Therefore the total product is divisible by (n!)(n-1)! 

 

So, (n!)!/(n!)(n-1)! is an integer.

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