Question

Show that the product of k consecutive natural numbers is divisible by k!​

Answer 1

Answer:

The product of  k  consecutive integers is divisible by  k! , in particular by  k  (provided  k≥1 ).

Step by step explanation:

If  m  is the first of the consecutive integers, set  n=m+k−1 , so

m(m+1)…(m+k−1)=n(n−1)…(n−k+1)  

so

(nk)=n(n−1)…(n−k+1)k!  

is the number of ways to choose  k  elements from  n , which is obviously an integer.

Answer 2

The product of k consecutive natural numbers. Hence, the product of k consecutive natural numbers is divisible by k!.

Let's consider k consecutive natural numbers starting from n, then the product of these numbers is:

n(n+1)(n+2)...(n+k-1)

Now, we need to prove that this product is divisible by k!.

We know that k! can be written as:

k! = k(k-1)(k-2)...(2)(1)

We can see that k! contains all the factors that appear in the product of k consecutive natural numbers.

To prove that the product of k consecutive natural numbers is divisible by k!, we need to show that each factor in k! appears at least once in the product of k consecutive natural numbers.

Let's consider the ith factor in k! which is (k-i+1). This factor appears in the product of k consecutive natural numbers in the following way:

(n+i-1)(n+i-2)...(n+1)n

We can see that the first factor in this expression is (n+i-1) which is the same as (k-i+1). Similarly, the second factor is (n+i-2) which is the same as (k-i+2). This pattern continues until the last factor which is n, and it is the same as k.

Therefore, each factor in k! appears at least once in the product of k consecutive natural numbers. Hence, the product of k consecutive natural numbers is divisible by k!.

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