How do you prove by induction that #n(n+1)(n+2)(n+3)# is divisible by #24# for any non-negative integer #n# ?

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Answer 1 · 2017-12-07T00:18:41

See explanation...

Let #P(n)# be the proposition:

#n(n+1)(n+2)(n+3)" "# is divisible by #24#

Base case

#P(0) = 0(1)(2)(3) = 0" "# is divisible by #24#

Induction step

Suppose #P(n)#.

Then:

#(n+1)(n+2)(n+3)(n+4)#

#= n(n+1)(n+2)(n+3)+4(n+1)(n+2)(n+3)#

We know that #n(n+1)(n+2)(n+3)# is divisible by #24#.

So we just have to show that #4(n+1)(n+2)(n+3)# is divisible by #24# too.

Note that exactly one of #(n+1)#, #(n+2)# and #(n+3)# will be divisible by #3#. So their product is also divisible by #3#.

Note that one of #(n+1)# and #(n+2)# is even. So their product is too.

Hence #(n+1)(n+2)(n+3)# is divisible by both #3# and #2# and therefore by their product #6#.

So #4(n+1)(n+2)(n+3)# is divisible by #4 * 6 = 24# as required.

Conclusion

Having shown that #P(0)# holds and #P(n) -> P(n+1)#, we can deduce that #P(n)# holds for all natural numbers #n#.