Question

For all positive integers $n,\frac{n^7}{7}+\frac{n^5}{5}+\frac{2n^3}{3}-\frac{n}{105}$ is a/an

• A. integer
• B. rational number
• C. real number
• D. none of these

Answer 1

Answer: (A)

Let $P\left(n\right)=\frac{n^7}{7}+\frac{n^5}{5}+\frac{2n^3}{3}-\frac{n}{105}$ is an integer.
For $n=1$
$P\left(1\right)\frac{{\left(1\right)}^7}{7}+\frac{{\left(1\right)}^5}{5}+\frac{2{\left(1\right)}^3}{3}-\frac{\left(1\right)}{105}$
$=\frac{1}{7}+\frac{1}{5}+\frac{2}{3}-\frac{1}{105}$
$=\frac{15+21+70-1}{105}=\frac{106-1}{105}$
$=\frac{105}{105}=1$
$\therefore P\left(1\right)$ is true.
Let $P\left(k\right)$ be true, then
$P\left(k\right)=\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$ is an integer
$\therefore \frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}=m,m\in Z...\left(i\right)$
For $n=k+1$, we have
$P\left(k+1\right)=\frac{{\left(k+1\right)}^7}{7}+\frac{{\left(k+1\right)}^5}{5}+\frac{2{\left(k+1\right)}^3}{3}-\frac{\left(k+1\right)}{105}$
$=\frac{1}{7}\left(k^7+7k^6+21k^5+35k^4+35k^3+21k^2+7k+1\right)$
$+\frac{1}{5}\left(k^5+5k^4+10k^3+10k^2+5k+1\right)$
$+\frac{2}{3}\left(k^3+3k^2+3k+1\right)-\frac{k}{105}-\frac{1}{105}$
$=\left(\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}\right)+k^6+3k^5+6k^4$
$+7k^3+7k^2+4k+1$
$=m+k^6+3k^5+6k^4+7k^3+7k^2+4k+1$[Using $\left(i\right)$]
= an integer
Thus $P\left(k\right)$ is true $\Rightarrow P\left(k+1\right)$ is true.
Hence by principle of mathematical induction, $P\left(n\right)$ is true
for all positive integers $n$.