Question

Show that 1n^ 3 + 2n + 3n ^2 is divisible by 2 and 3 for all positive integers n.

Answer 1

Prove:

Using mathemetical induction:

P(n) = $n^{3}+2n+3n^{2}$

for n=1

P(n)  = $1^{3}+2(1)+3(1)^{2}$ = 6

It is divisible by 2 and 3

Now, for n=k, $n > 0$

P(k) = $k^{3}+2k+3k^{2}$

Assuming P(k) is divisible by 2 and 3:

Now, for n=k+1:

P(k+1) = $(k+1)^{3}+2(k+1)+3(k+1)^{2}$

P(k+1) = $k^{3}+3k^{2}+3k+1+2k+2+3k^{2}+6k+3$

P(k+1) = $P(k)+3(k^{2}+3k+2)$

Since, we assumed that P(k) is divisible by 2 and 3, therefore, P(k+1) is also

divisible by 2 and 3.

Hence, by mathematical induction, P(n) = $n^{3}+2n+3n^{2}$ is divisible by 2 and 3 for all positive integer n.