Question

Prove by the Principle of Mathematical Induction: \(n (n^2 +5)\) is divisible by 6, for each natural number n.

Answer 1

Assume given statement

Let \(P(n)=n(n^2+5)\) is divisible by 6, for each natural number n.

Check that statement is true for \(n=1\)

\(P(1)=1(1^2+5)=6\)

\(\Rightarrow P(1)\) is divisible by 6

So, P(n) is true for \(n = 1\)

Assume P(k) to be true and then prove \(P(k+1)\) is true.

Lets assume P(k) is true So, P(k) is divisible by 6.

\(\Rightarrow k(k^2+5)=6q …(1)\)

\(P(k+1)=(k+1)[(k+1)^2+5]\)

\( =(k+1)[k^2+2k+1+5]\)

\( =(k+1)(k^2+2k+6) \)

\(=k^3+2k^2+6k+k^2+2k+6\)

\( =k^3+3k^2+8k+6\)

\( =k^3+5k+3k^2+3k+6 \)

\(=k(k^2+5)+3(k^2+k)+6\)

\( P(k+1)=6q+3k(k+1)+6 \)

Now, if k is odd \(k+1\) will be even and if k is even \(k+1\) will be odd and we know that even multiplied with odd is even.

So, \(k(k+1)\) is always even

\(\Rightarrow k(k+1)=2p\)

\( \Rightarrow P(k+1)=6q+3.2p+6 \)

\(=6q+6p+6 =6(q+p+1)\)

\( \therefore P(k+1)\) is divisible by 6.

Hence, \(P(k+1)\) is true whenever P(k) is true.

Hence, By Principle of mathematical Induction P(n) is true for all natural numbers n.