We can solve this problem using statement of principle of mathematical induction.
Statement of the Principle of Mathematical Induction:
Let P (n) be a statement involving the natural number ‘n’
such that
a. (1) is true i.e. P(n) is true for n = 1 and (m+1) is true, whenever P(m) is true
i.e. P(m) is true ⇒ P(m+1) is true.
Then, P (n) is true for all natural numbers n.
By using these steps we can find the answer
For n = 1
r+1 = 1! (r+1)
We shall now to prove (r+1) (r+2) ..... (r+n) is divisible by n!.
Let P(n) : (r+1) (r+2) ..... (r+n) = n! . k
where k is an integer.
when n = 1,r+1 = 1!.(r+1)
Therefore, P(1) is true.
Let P(m) be true. That is,
(r+1)(r+2) (r+3) ..... (r+m) = m! × k ....(1)
Now,
(r + 1) (r + 2) (r + 3) ... (r + m)
(r+) = r (r + 1) (r + 2) (r + 3) ... (r + m) + (m + 1) (r + 1) (r + 2) (r + 3) ... (r + m)
=
+ (m + 1) (m!) k, using (i)
= (m + 1)! ×
+ (m + 1)! × k
= (m+1)!
(Cr−1+k) = (m+1)! (integer + k)
= (m+1)!.
k1, where k1 is an integer.
Therefore, P(m+1) is true.
Thus P(m) is true ⇒ P(m+1) is true.
Therefore, P(n) is true for all n∈N.
Hence, the greatest positive integer which divides (r+1)
(r+2) (r+3) .... (r+n) for all n ∈ N is n!.