Question

. Use mathematical induction to show that the product of any n consecutive positive integers is divisible by n!.

Hint: Use the identity
m(m + 1)···(m + n 1)/n!=(m 1)m(m + 1)···(m + n 2)/n! + m(m + 1)···(m + n 2)/(n 1)!

Answer 1

Mathematical induction is a proof technique where we establish a base case and then show that if the statement holds for an arbitrary value, it also holds for the next value. In this case, we want to prove that the product of any
`\(n\)` consecutive positive integers is divisible by
`\(n!\)`.

Base Case
`(n=1)`: When
`\(n=1\)`, the product of a single positive integer is itself, and
`\(1!\)` is also 1. Therefore, the base case holds true.

Inductive Step: Assume that the product of
`k` consecutive positive integers is divisible by
`\(k!\)` for some arbitrary
`k`.

Now, we want to show that the product of
`\((k+1)\)` consecutive positive integers is divisible by
`\((k+1)!\)`. We use the provided identity:

`\[ (m(m + 1) \cdots (m + k))/(k!) = ((m-1)m(m + 1) \cdots (m + k - 1))/((k-1)!) + (m(m + 1) \cdots (m + k - 1))/(k!) \]`

This identity allows us to express the product of
`\((k+1)\)` consecutive integers as the sum of two terms, each of which is divisible by
`\(k!\)`.

By the inductive assumption, both terms are divisible by
`\(k!\)`, and hence their sum, representing the product of
`\((k+1)\)` consecutive positive integers, is also divisible by
`\((k+1)!\)`.

Thus, by mathematical induction, we have shown that the product of any
`\(n\)` consecutive positive integers is divisible by
`\(n!\)`.