# Prove that n(n+1)(2n+1) is divisible by 6 for all n?N by using principle of mathematical induction.

Let the given statement be $P\left(n\right)=n\left(n+1\right)\left(2n+1\right)$ which is divisible by 6

For $n=1,P\left(1\right):1\times \left(1+1\right)\left(2+1\right)=6$, which is true

Assume that $P\left(k\right)$ is true for some positive integers

*k*i.e.,

*P*(

*k*) is divisible by 6

We shall now prove that

*P*(

*k +*1) is also true

$P(k+1)=\left(k+1\right)\left(k+2\right)\left[2\left(k+1\right)+1\right]=\left(k+1\right)\left(k+2\right)\left[2k+3\right]=\left(k+1\right)\left(k+2\right)\left[2k+1+2\right]\phantom{\rule{0ex}{0ex}}=k\left(k+1\right)\left[\left(2k+1\right)+2\right]+2\left(k+1\right)\left[\left(2k+1\right)+2\right]\phantom{\rule{0ex}{0ex}}=k(k+1)(2k+1)+2k(k+1)+2(k+1)(2k+1)+4(k+1)\phantom{\rule{0ex}{0ex}}=k(k+1)(2k+1)+2(k+1)\left[k+2k+1+2\right]\phantom{\rule{0ex}{0ex}}=k(k+1)(2k+1)+2(k+1)\left[3k+3\right]=k(k+1)(2k+1)+6(k+1)\left(k+1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

First term is divisible by 6 as we assumed, and second term is also divisible by 6.

Thus,

*P*(

*k*+ 1) is divisible by 6, whenever

*P*(

*k*) is true.

Hence, from the principle of mathematical induction

*P*(

*n*) is true for $n\in N$

Regards

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