Question

Mathematical induction question

Answer 1

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

Answer 2

I'be done mathematical induction before but the questions aren't usually phrased like this.

Answer 3

wat do u think

Answer 4

Well all the other times I did mathematical induction it wasn't to prove something like this.

Answer 5

Does this look like :
(n^3 + 2n)/3 = n?

Answer 6

Usually it looks like this for me
4 * 6 + 5 * 7 + 6 * 8 + ... + 4n( 4n + 2) = ( 4(4n+1)(8n+7) )/6

Answer 7

I can do those easily

Answer 8

\(\large \begin{align} \color{black}{n^3+2n \hspace{.33em}\\~\\
\normalsize \text{for case 1, n=1 } \hspace{.33em}\\~\\
n^3+2n \hspace{.33em}\\~\\
1^3+2(1) \hspace{.33em}\\~\\
=3 \hspace{.33em}\\~\\
\normalsize \text{so its divisible by 3 } \hspace{.33em}\\~\\
\normalsize \text{for case 2, let n=k } \hspace{.33em}\\~\\
\normalsize \text{assume that it is true for } \hspace{.33em}\\~\\
\color{red }{k^3+2k}\hspace{.33em}\\~\\
\normalsize \text{you have to prove that it is also true for } \hspace{.33em}\\~\\
\color{green}{(k+1)^3+2(k+1)} \hspace{.33em}\\~\\
\normalsize \text{so } \hspace{.33em}\\~\\
(k+1)^3+2(k+1) \hspace{.33em}\\~\\
=(k+1)(k^2-k+1)+2k+2 \hspace{.33em}\\~\\
=k^3-k^2+k+k^2-k+1+2k+2 \hspace{.33em}\\~\\
= \color{red}{(k^3+2k)}+(3) \hspace{.33em}\\~\\
\normalsize \text{you assumed that the red part is divisible by 3 } \hspace{.33em}\\~\\
\normalsize \text{and the remaining is divisible by 3 } \hspace{.33em}\\~\\
}\end{align}\)

Answer 9

So I dont set it to anything until I plug in 1 and see what I get?

Answer 10

in mathmatical induction we there are only two case , and we plug only 1 and other variable for example k to prove it

Answer 11

http://www.mathsisfun.com/algebra/mathematical-induction.html

Answer 12

and also we plug (k+1) to prove that it is also prove for the next consecutive number

Answer 13

Ok.