$\frac{n^{7}}{7}+\frac{n^{5}}{5}+\frac{2 n^{3}}{3}-\frac{n}{105}$

Question

Professor PC Thomas Classes · Answer

We prove the result by principle of mathematical induction. Let $P(n)=\frac{n^{7}}{7}+\frac{n^{5}}{5}+\frac{2 n^{3}}{3}-\frac{n}{105}, n \in N$\ Now $P(1)=\frac{1}{7}+\frac{1}{5}+\frac{2}{3}-\frac{1}{105}=1$, an integer\ $	herefore \quad$ The result is true for $n=1$\ Assume that it is true for $n=k$\ i.e. $P(k)=\frac{k^{7}}{7}+\frac{k^{5}}{5}+\frac{2 k^{3}}{3}-\frac{k}{105}$ is an\ integer, say $l$\ For $n=k+1$,\ $P(k+1)=\frac{(k+1)^{7}}{7}+\frac{(k+1)^{5}}{5}+\frac{2(k+1)^{3}}{3}-\frac{(k+1)}{105}$\ $=\frac{1}{7}\left(k^{7}+7 k^{6}+21 k^{5}+35 k^{4}ight.$\ $\left.+35 k^{3}+21 k^{2}+7 k+1ight)$\ $+\frac{1}{5}\left(k^{5}+5 k^{4}+10 k^{3}+10 k^{2}+5 k+1ight)$\ $+\frac{2}{3}\left(k^{3}+3 k^{2}+3 k+1ight)-\frac{k}{105}-\frac{1}{105}$\ $=\left(\frac{k^{7}}{7}+\frac{k^{5}}{5}+\frac{2 k^{3}}{3}-\frac{k}{105}ight)$\ $+\left(k^{6}+3 k^{5}+6 k^{4}+7 k^{3}+7 k^{2}+4 kight)$\ $+\left(\frac{1}{7}+\frac{1}{5}+\frac{2}{3}-\frac{2}{105}ight)$\ $=l+$ some integer $+1=$ an integer\ Hence by the principle of mathematical induction, the result is true for all $n \in N$.\