Question

Prove by the principle of mathematical induction that $\frac{n^5}{5}+\frac{n^3}{3}+\frac{7n}{15}$ is a natural number for all n $ϵ$ N.

Answer 1

Let P(n) be the statement given by

P(n) : $\frac{n^5}{5}+\frac{n^3}{3}+\frac{7n}{15}$ is a natural number.

For n = 1.

P(1) : $\frac{1^5}{5}+\frac{1^3}{3}+\frac{7\times 1}{15}=\frac{1}{5}+\frac{1}{3}+\frac{7}{15}$

$=\frac{3+5+7}{15}=\frac{15}{15}=1$

which is a natural number.

Hence, P(1) is true.

Assume that, P(n) is true for some natural number k.

$\therefore P\left(k\right)=\frac{k^5}{5}+\frac{k^3}{5}+\frac{7k}{15}$ is a natural number.

Now, we will show that P(k + 1) is true whenever P(k) is true for which we have to show that

$\frac{(k+1{)}^5}{5}+\frac{(k+1{)}^3}{3}+\frac{7(k+1)}{15}$ is a natural number.

Now, $\frac{(k+1{)}^5}{5}+\frac{(k+1{)}^3}{3}+\frac{7(k+1)}{15}$

$=\frac{1}{5}(k^5+5k^4+10k^3+10k^2+5k+1)+\frac{1}{3}[k^3+3k^2+3k+1]+\frac{7}{15}[k+1]$

$\left[\begin{array}{c}\because (1+k{)}^5={}^5{C}_0{k}^5+{}^5{C}_1{k}^4+{}^5{C}_2{k}^3+{}^5{C}_3{k}^2\\ +{}^5{C}_{4}k+{}^5{C}_5=k^5+5k^4+10k^3+10k^2+5k+1\end{array}\right]$

$=[\frac{k^5}{5}+\frac{k^3}{3}+\frac{7k}{15}]+(k^4+2k^3+3k^2+2k)$

$+\frac{1}{5}+\frac{1}{3}+\frac{7}{15}$

$=\lambda +k^4+2k^3+3k^2+2k+1$

which is natural number.

$\therefore P(k+1)$ is true.

Hence, by principle of mathematical induction, P(n) is true, $\forall nϵN$