Question

If r is a fixed positive integer, prove by induction that (r+1)(r+2)(r+3)....(r+n) is divisible by n!

• A. I want to see the solution
• B. Take me to next question

Answer 1

The correct option is B

Take me to next question

Let P(n): (r+1)(r+2)(r+3)....(r+n)=n!.k,

Where k is an integer.

When n=1,r+1=1!.(r+1)

$\therefore P\left(1\right)$ is true.

let P(m) be true,i.e.,

(r+1)(r+2)(r+3)....(r+m)=m!.k ....(1)

Now, (r+1)(r+2)(r+3)....(r+m)(r+m+1)

= r(r+1)(r+2)(r+3)....(r+m)+(m+1)(r+1)(r+2)(r+3)...(r+m)

$=\frac{(r+m)!}{(r-1)!}+(m+1)(m!)k,using\left(1\right)$

=(m+1)!.$\frac{(r+m)!}{(r-1)!(m+1)!}+(m+1)!.k$

=(m+1)!.${(}^{r+m}{C}_{r-1}+k)$

=(m+1)!(integer+k)

P(m+1) is true;
Thus, p (m) is true
P (m+1) is true:

Thus,P(m) is true $\Rightarrow$P(m+1)is true.

P(n) is true for all $nϵN$.