BITSAT 2023 Shift 1 Paper

Let the statement $P(n)$ be defined as$P(n): \frac{n^5}{5} + \frac{n^3}{3} + \frac{7n}{15}$ is a natural number for all $n \in N$.Step I : For $n=1, P(1): \frac{1}{5}+\frac{1}{3}+\frac{7}{15}=1 \in N$Hence, it is true for $n=1$.Step II : Let it is true for $n=k$,i.e. $\frac{k^5}{5}+\frac{k^3}{3}+\frac{7k}{15}=\lambda \in N$ . . . (i) $\text{ Step III : For } n=k+1$$\frac{(k+1)^5}{5}+\frac{(k+1)^3}{3}+\frac{7(k+1)}{15}$$= \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 + \frac{7}{15}$$= \left(\frac{k^5}{5}+\frac{k^3}{3}+\frac{7}{15}k\right) + (k^4+2k^3+3k^2+2k)$$+ \frac{1}{5}+\frac{1}{3}+\frac{7}{15}$$=\lambda+ k^4+2k^3+3k^2+2k+1$[using equation (i)] which is a natural number, since $\lambda k \in N$. Therefore, $P(k+1)$ is true, when $P(k)$ is true, Hence, from the principle of mathematical induction, the statement is true for all natural numbers $n$.