Answer to Question #200204 in Real Analysis for Rajkumar

Question

Using the principle of mathematical induction, prove that

(n⁵/5) + (n³/3) + (7n/15) is a natural number, ∀ n ∈ N .

Answer 1

Let $P(n)=\dfrac{n^{5}}{5}+\dfrac{n^{3}}{3}+\dfrac{7n}{15}$ is a natural number for all $n$ $∈N$

now putting $n=1$ , we get -

$P(1)=\dfrac{1^{5}}{5}+\dfrac{1^{3}}{3}+\dfrac{7\times1}{15}=\dfrac{15}{15}=1$ .

which is a natural number .

Hence , P(1) is true .

Let us prove that P(n) is true for some natural number n=k.

$\therefore$ $P(k)=\dfrac{k^{5}}{5}+\dfrac{k^{3}}{3}+\dfrac{7k}{15}$ is a natural number ..........(1)

Now we have to prove that $P(k+1)$ is true.

$P(k+1)=$ $\dfrac{(k+1)^{5}}{5}+\dfrac{(k+1)^{3}}{3}+\dfrac{7(k+1)}{15}$

$=$ $\dfrac{k^{5}+5k^{4}+10k^{3}+10k^{2}+5k+1}{5}+\dfrac{k^{3}+1+3k^{2}+3k}{3}+\dfrac{7k+7}{15}$

$=\dfrac{k^{5}}{5}+\dfrac{k^{3}}{3}+\dfrac{7k}{15}+$ $\dfrac{5k^{4}+10k^{3}+10k^{2}+5k+1}{5}+$ $\dfrac{1+3k^{2}+3k}{3}+\dfrac{7}{15}$

$=$ $\dfrac{k^{5}}{5}+\dfrac{k^{3}}{3}+\dfrac{7k}{15} \ +k^{4}+2k^{3}+2k^{2}+k+k^{2}+k+\dfrac{1}{5}+\dfrac{1}{3}+\dfrac{7}{15}$

$=$ $\dfrac{k^{5}}{5}+\dfrac{k^{3}}{3}+\dfrac{7k}{15}+k^{4}+2k^{3}+3k^{2}+2k+1$

which is a natural number .....by using .....(1)

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

so by principal of mathematical induction

$P(n)$ is true for any natural number n .