Proof using mathematical induction

[cdummie]

Prove that n^5 - 5n^3 + 4n is divisible by 120. for every natural number n greater or equal to 3.

First, i checked if it works for n=3 and it does,

so i could assume it works for some k>=3 so i could write k^5 - 5k^3 + 4k as 120*a a is natural number

so for k+1 i have:

(k+1)^5 - 5(k+1)^3 + 4(k+1)

and after applying Binomial theorem i got:

k^5 - 5k^3 + 4k + 5k^4 + 10k^3 - 5k^2 - 10k =

120a + 5k^4 + 10k^3 - 5k^2 - 10k

so now i only have to prove that 5k^4 + 10k^3 - 5k^2 - 10k is divisible by 120 but the problem is that i don't know how, doing induction again would be too complicated, so i tried to see what happen if k is even and what happens if k is odd number, if it's even i can write it as 2 times some number b but i end up with

80b^4 + 80b^3 - 20b^2 - 20b which is divisible by 20 and that doesn't mean is that is divisible by 120. What can i do, is this good approach?

 

[PeroK]

Prove that n^5 - 5n^3 + 4n is divisible by 120. for every natural number n greater or equal to 3.

First, i checked if it works for n=3 and it does,

so i could assume it works for some k>=3 so i could write k^5 - 5k^3 + 4k as 120*a a is natural number

so for k+1 i have:

(k+1)^5 - 5(k+1)^3 + 4(k+1)

and after applying Binomial theorem i got:

k^5 - 5k^3 + 4k + 5k^4 + 10k^3 - 5k^2 - 10k =

120a + 5k^4 + 10k^3 - 5k^2 - 10k

so now i only have to prove that 5k^4 + 10k^3 - 5k^2 - 10k is divisible by 120 but the problem is that i don't know how, doing induction again would be too complicated, so i tried to see what happen if k is even and what happens if k is odd number, if it's even i can write it as 2 times some number b but i end up with

80b^4 + 80b^3 - 20b^2 - 20b which is divisible by 20 and that doesn't mean is that is divisible by 120. What can i do, is this good approach?

Re another approach: what about factorisation?

 

[cdummie]

Re another approach: what about factorisation?

I will have product of five consecutive natural numbers which is divisible by 120, but i have to use induction to prove this.

 

[mathman]

I will have product of five consecutive natural numbers which is divisible by 120, but i have to use induction to prove this.

You don't need induction. 120=2.3.4.5. 5 consecutive positive numbers are divisible by 2, 3, 4, and 5. To clarify about even numbers - there will be 2 even numbers in the string, one of which will be divisible by 4.

 

[Mark44]

5 consecutive positive numbers are divisible by 2, 3, 4, and 5.

It would be useful (and maybe even required) for the OP to prove this statement.

 

[cdummie]

You don't need induction. 120=2.3.4.5. 5 consecutive positive numbers are divisible by 2, 3, 4, and 5. To clarify about even numbers - there will be 2 even numbers in the string, one of which will be divisible by 4.

OK, i don't need induction to prove this, there are different ways, i know. But the problem statement is such that i have to use induction to prove it, otherwise it doesn't count, so how can i solve it using induction?

 

[PeroK]

OK, i don't need induction to prove this, there are different ways, i know. But the problem statement is such that i have to use induction to prove it, otherwise it doesn't count, so how can i solve it using induction?

You rejected the idea of multiple inductions for no good reason that I can see. If you must use induction, go back to your original approach and work on the second polynomial you generated. It's got a factor of 5, which should make things easier.

 

[mathman]

$5k^4+10k^3-5k^2-10k=5(k-1)k(k+1)(k+2)$. You can work it out from here.

 

[cdummie]

$5k^4+10k^3-5k^2-10k=5(k-1)k(k+1)(k+2)$. You can work it out from here.

Actually i tried using induction more than once and i ended up with doing induction five times but i solved it, thanks for help.

 

[mathwonk]

maybe technically induction is required, but isn't it obvious that in any sequence of 5 consecutive natural numbers, at least one is divisible by 5, one is divisible by 3 and one is divisible by 4, and another is divisible by 2? bingo. but i guess the point is to practice induction.

 

[FactChecker]

$5k^4+10k^3-5k^2-10k=5(k-1)k(k+1)(k+2)$. You can work it out from here.

Factoring would also have worked for the first step. n⁵ - 5n³ + 4n = (n-2)(n-1)n(n+1)(n+2)

 

[cdummie]

maybe technically induction is required, but isn't it obvious that in any sequence of 5 consecutive natural numbers, at least one is divisible by 5, one is divisible by 3 and one is divisible by 4, and another is divisible by 2? bingo. but i guess the point is to practice induction.

Exactly, that is the point. Practicing mathematical induction.