 # Use mathematical induction to prove that n(n2+5) is divisible by 6 for n∈N fortdefruitI 2020-11-09 Answered

Use mathematical induction to prove that $n\left({n}^{2}+5\right)$ is divisible by 6 for $n\in N$

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Let P(n) denotes that $n\left({n}^{n+5}\right)$ is divisible by 6, for each natural number n.
$P\left(1\right):1\left({1}^{2+5}\right)=6$, which is divisible by 6. Therefore the result is true for n=1.
Now, assume that P(n) is true for n=k. That is for some m∈N. We have to prove that P(k+1) is also true.
$P\left(k+1\right):\left(k+1\right)\left[{\left(k+1\right)}^{2+5}\right]$

$=\left(k+1\right)\left[{k}^{2}+2k+1+5\right]$

$=\left(k+1\right)\left[{k}^{2}+2k+6\right]$

$={k}^{3}+2{k}^{2}+6k+{k}^{2}\cdot 2k+6$

$={k}^{3}+3{k}^{2}+8k+6$

$={k}^{3}+5k+3{k}^{2}+3k+6$

$=k\left({k}^{2}+5\right)+3\left({k}^{2}+k+2\right)$

$=\left(6m\right)+3\left({k}^{2}+k+2\right)$
Since ${k}^{2}+k+2$ is divisible by 22 for every even and odd k. Therefore $3\left({k}^{2}+k+2\right)$ is divisible by 66 and hence $\left(6m\right)+3\left({k}^{2}+k+2\right)$ is divisible by 6. Therefore, $P\left(k+1\right)$ is true whenever P(k) is true. Hence, by the principle of mathematical induction P(n) is true.