Question

show that 4n(n^2-1) is divisible by 24
show that 4n(n^2-1) is divisible by 24
@Mathematics

Answer 1

my approach is as follows:
n(n^2-1) is divisible by 6

clearly true for n=2 as that is 2(4-1) = 6
now assume true for n, and try and prove for n+1
expand to n(n-1)(n+1) for n
and for n+1 its
n(n+1)(n+2)
which is n(n+1)(n-1 +3)
which is n(n+1)(n-1) + 3n(n+1)
now lhs is divisible by 6 as it is the case for n
and rhs is an even multiple of 3 which is always divisble by 6,
its even because n (n+1) must be an even number

now is there a better way than that?

Answer 2

induction works

Answer 3

except for the fact that 6 is not divisible by 24!

Answer 4

should say
\[n\geq 3\]

Answer 5

simplify and work the smaller numbers

Answer 6

yep

Answer 7

i'm saying that 4 x z/24 is like saying z/6

Answer 8

funny the question didn't say n>= 3

Answer 9

4n(n^2-1); n = 0,1,2,3,...

{0,0,24,96, ...}

Answer 10

yeah i know, i just meant that the whole thing should start with
\[n\geq 3\] that is all

Answer 11

is there a better way to show it than mine

Answer 12

is n across the reals or the integers?

Answer 13

integers only

Answer 14

you have the right idea. you want simply that
\[(n-1)n(n+1)\] is divisible by 6, but i doubt you need induction.

Answer 15

you have three consecutive integers. one is even and one is divisible by 3.

Answer 16

good point

Answer 17

thanks all for your help