For every natural number n, 3|(n^3 + 5n + 6)?

Question

I'm under the assumption that it's true but I can't prove so by induction.

Answer 1

The answer is YES!

Consider ANY natural integer number 3k + x, where x = -1, 0, 1. These are three successive integers for a given ' k,' and so that covers all the integers.

We can ignore the ' 6,' since that is divisible by 3 already.

Then n^3 + 5n = (3k + x)^3 + 5(3k + x) =

(27k^3 + 27xk^2 + 9kx^2 + x^3) + (15k + 5x). Ignoring the terms with a coefficient already a multiple of 3, we are left with just :

x^3 + 5x.

Now evaluate this for x = -1, 0, and 1. We obtain, successively,

-6, 0, and 6.

The ' 0 ' is redundant, and +/- 6 are both divisible by 3.

Hence ALL numbers of this form ARE divisible by 3 !

QED

Live long and prosper.

Answer 2

It is true. Here's how to simplify the work a bit:
First, since 6 is already a multiple of 3, it's
enough to shows 3|n³+5n for all n.
Next, n³+5n = n(n²+5).
If n is a multiple of 3, we are done.
Otherwise, we have to show n²+5 is a
multiple of 3 for n = 3m +1 and n = 3m+2.
If n = 3m+1, (3m+1)² + 5 = 9m²+6m+6 = 3(3m²+2m+2),
so it's a multiple of 3.
If n =3m+2, (3m+2)² + 5 = 9m² + 12m +9
= 3(3m²+4m+3), also a multiple of 3.
So the result holds for all n.

Answer 3

n^3+ 5n + 6 = n(n^2+5) + 6
= n[(n+1)*(n+2) -3n + 3) + 6
= n(n+1)(n+2) -3n(n-1) + 6
n(n+1)(n+2) is divisible by 3 as product of 3 consecutive numbers and rest are also
so is the expression.

Answer 4

All natural numbers can can be expressed in the forms:
3k, 3k+1, or 3k+2. Just try plugging in each of the possibilities, and seeing if a 3 factors out.

Answer 5

utilising induction tutor for the organic numbers n a million^2+2^2+3^2+...n^2=(n(n+a million)(2n+a million))/6 pf if n=a million then for sure a million^2 = (a million)(2)(3)/6 =a million so the assertion is real whilst n=a million. assume assertion is real for n=ok. it is we assume a million^2+2^2+3^2+...ok^2=(ok(ok+a million)(2k+a million))/6 Now, we tutor the assertion is real for n=ok+a million. a million^2+2^2+3^2+...+ok^2+(ok+a million)^2 by skill of induction speculation all of us comprehend that a million^2+2^2+3^2+...+ok^2 = (ok(ok+a million)(2k+a million))/6 if we upload (ok+a million)^2 to the two sides we get: a million^2+2^2+3^2+...+ok^2 + (ok+a million)^2 = (ok(ok+a million)(2k+a million))/6 + (ok+a million)^2 Now we simplify the main superb hand section. (ok(ok+a million)(2k+a million))/6 + (ok+a million)^2 = ( ok(ok+a million)(2k+a million))/6 + 6(ok+a million)^2/6= ( ok(ok+a million)(2k+a million) + 6(ok+a million)^2 ) /6= [ (ok+a million)(ok(2k+a million) + 6(ok+a million) ) ]/6= [(ok+a million)(2k^2 +7k +6) ]/6= [(ok+a million)(2k + 3)(ok + 2) ]/6= [(ok+a million)(2(ok+a million)+a million)((ok+a million)+a million)]/6 this shows the assertion is real for n =ok+a million so the assertion is real for all n by skill of induction.