Verified Answer :

apologiabiology
  1. prove it is true for n=1
    2. assume n=k
    3. prove that n=k+1 is true as well


    so

    1.
    \(\frac{n^3+2n}{3}\)=
    \(\frac{1^3+2(1)}{3}\)=
    \(\frac{1+2}{3}\)=1
    we got a whole number, true


    2.
    \(\frac{k^3+2k}{3}\)
    if everything clears, then it is divisble


    3.
    \(\frac{(k+1)^3+2(k+1)}{3}\) =
    \(\frac{(k+1)^3+2(k+1)}{3}\) =
    \(\frac{k^3+3k^2+3k+1+2k+2)}{3}\)=
    \(\frac{k^3+3k^2+5k+3)}{3}\)
    we know that if z is divisble by 3, then z+3 is divisble b 3
    also, 3k/3=a whole number when k= a whole number

    \(\frac{k^3+2k}{3} + \frac{3k^2+3k+3}{3}\)=
    \(\frac{k^3+2k}{3} + k^2+k+1\)=
    since the k²+k+1 part cleared, it is divisble by 3

    we found that it simplified back to \(\frac{k^3+2k}{3}\)

    done

Answer:

We have to use the mathematical induction to  prove the statement is true for all positive integers n.

The integer \(n^3+2n\) is divisible by 3 for every positive integer n.

  • for n=1

\(n^3+2n=1+2=3\) is divisible by 3.

Hence, the statement holds true for n=1.

  • Let us assume that the statement holds true for n=k.

i.e. \(k^3+2k\) is divisible by 3.---------(2)

  • Now we will prove that the statement is true for n=k+1.

i.e. \((k+1)^3+2(k+1)\) is divisible by 3.

We know that:

\((k+1)^3=k^3+1+3k^2+3k\)

and \(2(k+1)=2k+2\)

Hence,

\((k+1)^3+2(k+1)=k^3+1+3k^2+3k+2k+2\\\\(k+1)^3+2(k+1)=(k^3+2k)+3k^2+3k+3=(k^3+2k)+3(k^2+k+1)\)

As we know that:

\((k^3+2k)\) was divisible as by using the second statement.

Also:

\(3(k^2+k+1)\) is divisible by 3.

Hence, the addition:

\((k^3+2k)+3(k^2+k+1)\) is divisible by 3.

Hence, the statement holds true for n=k+1.

Hence by the mathematical induction it is proved that:

The integer \(n^3+2n\) is divisible by 3 for every positive integer n.

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