Question

Prove that 2n^3 + 3n^2 + n is divisible by 6 for every integer n > 1 by mathematical induction. (7 marks)

Answer 1

Answer:

The answer is True

Step-by-step explanation:

A mathematical induction consists in only 2 steps:  

First step: Show the proposition is true for the first one valid integer number.

Second step: Show that if any one is true then the next one is true  

Finally, if first step and second step are true, then the complete proposition is true.

So, given $2*n^3+3*n^2+n$

First step: using and replacing n=2 (the first valid integer number >1)

$2*(2)^3 +3*(2)^2+2=30$

$\frac{30}{6} =5$

As the result is an integer number, so the first step is true.

Second step: using any next number, $n+1$, let it replace

$2*(n+1)^3+3*(n+1)^2+(n+1)\\2*(n^3+3*n^2+3*n+1)+3*(n^2+2*n+1)^2+(n+1)\\2*n^3+6*n^2+6*n+2+3*n^2+6*n+3+n+1\\(2*n^3+3*n^2+n)+(6*n^2+12*n+6)$

As the First step is true, we know that

$2*n^3+3*n^2+n$$=6*k$,

So let it replace in the previous expression

$6*k+6*(n^2+2*n+1)\\6*[k+(n^2+2*n+1)]$

Finally

$\frac{6*[k+(n^2+2*n+1)]}{6} =k+(n^2+2*n+1)$

where the last expression is an integer number

So the second step is true, and the complete proposition is True