How to prove this, using mathematical induction

How do I prove that n^3+2n is divisible by 3 for all positive integers of n, using mathematical induction?

Thanks [:

For n = 1
1^3 + 2*1 =3
is divisible of 3

Let say for n=k this is true
then k^3 + 2k = p(say)

now for n = k +1
(k +1)^3 + 2(k +1)
= k^3 + 3k^2 + 3k+ 1 + 2k +2
arrange this as
k^3 + 2k + 3k^2 + 3k + 3
= (k^3 + 2k) + 3(k^2 + k +1)
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=p 3(x) say
p is divisible by 3(assumption)
and 3 into x is also divisible of three(fact)

So p+ 3(x) is divisible by 3

So (k + 1)^3 + 2(k+1) is divisible of three

So for n = 1 it is true
if n = k it is rue then for n=k+1 also it is true
So by Mathematical Induction it is true for all positive n.



Regards
Muza

1.

Assume that it is true for some value of n; then

n³ + 2n = 3k for some value of k.

On this assumption,

2.

Prove that it is true for (n + 1) :

(n + 1)³ + 2(n + 1) = n³ + 3n² + 3n + 1 + 2n + 2

. . . . . . . . . . . . . . . = n³ + 3n² + 5n + 3

. . . . . . . . . . . . . . . = (n³ + 2n) + 3n² + 3

. . . . . . . . . . . . . . . = (n³ + 2n) + 3(n² + 1)

. . . . . . . . . . . . . . . = 3k + 3(n² + 1)

. . . . . . . . . . . . . . . = 3(k + n² + 1)

which contains a factor of 3, therefore the expression is divisible by 3.


3.

Show that n³ + 2n is divisible for a specific value of n ( for n = 1, preferably)

n³ + 2n = 1³ + 2(1) = 3

i.e. when n = 1, n³ + 2n is divisible by 3.

Therefore (1 + 1)³ + 2(1 + 1) is also divisible by 3

and so on.

Q E D

Let n=1 , 1+2=3 div by 3 , assume true for n=k ie k^3+2k is div by 3 , now prove tru 4 n=k+1 ie (k+1)^3+2(k+1) = k^3+2k+3(k^2+k+1) , assumed that k^3+2k is div by 3 and the other expression is a multiple of 3 , end