Question

The integer n3 + 2n is divisible by 3 for every positive integer n

prove it by math induction

is it my proof right ?

Answer 1

n(n^2 +2)/3

for n=1 ---- 1(1+2)=3/3 =1
n=2 ---- 2(4+2)=12/3=4
n=3 ---- 3(9+2)=33/3=11

for n ---- n(n^2 +2) /3

so using mathinduction we write it for n=k and suppos it true
what will be k(k^2 +2)/3 suppos it true

so than for k=k+1 we get

(k+1)((k+1)^2 +2) /3

for k=1 --- (1+1)((1+1)^2 +2) /3
2(2^2 +2) /3 = 2(4+2) /3 =2*6 /3 = 12/3 =4

Answer 2

prove it true for n = k which you have done

then assume its true for n = k

then let \[k(k^2 + 2) = 3p\]

which is really saying that the kth them has factors of 3 and some number p

prove n = k + 1

so \[(k + 1)((k + 1)^2 + 2) = 3q\]

for some integer q

then LHS

\[(k+1)(k^2 + 2k + 1 +2) \]

simplified its

\[k^3 +3k^2 + 5k + 3 \]

which can be written as

\[k^3 + 2k + 3k^2 + 3k + 3 = 3p + 3(k^2 + k + 1)\]

now the common factor 3 can be removed so that

\[3(p + k^2 + k + 1) = 3q \]

therefore left hand side = right hand side.... and \[q = p + k^2 + k + 1\]

therefore its true for n = k + 1

hope this helps

Answer 3

thank you very nice

Answer 4

we don't put k =1 when proving in math induction

Answer 5

@aajugdar yes you do. you start with k=1
then assume for k,
then check for k+1

Answer 6

well in here its n=1
n=k
and n =k+1 ne
I was correcting his equation solving method :D
in n=k+1 he did put k=1 again

Answer 7

@jhonyy9 still confused?

Answer 8

so thank you for your opinion

but this k=1 after the k=k+1 just wann being an example that is true too ,the same for k=1

hope that you will anderstand me

thank you

bye

Answer 9

Who needs induction? Let's just take cases!
Any positive integer would be divisible by 3, or would leave a remainder of 1 or 2 when divided by 3 :)

For any positive integer n, we have...
\[\Large n^3+2n=n(n^2+2)\]
So, if n is divisible by 3, then clearly, the proof is done :D
Suppose n leaves a remainder of 1, when divided by 3. Then there's an integer h, such that...
\[\Large n=3h+1\]\[\Large n(n^2+2)=n[(3h+1)^2+2]=n(9h^2+6h+1+2)\]\[\Large =3n(3h^2+2h+1)\]So it's still divisible by 3.


Suppose instead, that n leaves a remainder of 2, when divided by 3. Then there's an integer k, such that
\[\Large n=3k+2\]\[\Large n(n^2+2)=n[(3k+2)^2+2]=n(9k^2+12k+4+2)\]\[\Large =3n(3k^2+4k+2)\]So it's still divisible by 3.


So in all cases, it will be divisible by 3 :)
Problem solved :>