Question

For all $n\in I^{+}$, the statement $P\left(n\right)=\frac{n^7}{7}+\frac{n^5}{5}+\frac{2}{3}{n}^3-\frac{n}{105}$ is a natural number is true, if

• A. $P\left(1\right)$ is true
• B. $P\left(2\right)$ is true
• C. $P\left(3\right)$ is true
• D. True $\forall n\in N$

Answer 1

Answer: (A), (B), (D)

$P\left(1\right)$ is true
$P\left(2\right)$ is true
True $\forall n\in N$

Given $P\left(n\right)=\frac{n^7}{7}+\frac{n^5}{5}+\frac{2}{3}{n}^3-\frac{n}{105}$
Put $n=1$ in the given expression
$\frac{1}{7}+\frac{1}{5}+\frac{2}{3}-\frac{1}{105}$
$=\frac{105}{105}=1$
which is a natural number.
Hence, P(1) is true.
Put $n=2$ in the given expression
$\frac{2^7}{7}+\frac{2^5}{5}+\frac{2^4}{3}-\frac{2}{105}$
$=\frac{3150}{105}=30$
which is a natural number.
Hence, P(2) is true.
Given that $P\left(n\right)$ is true
$\frac{n^7}{7}+\frac{n^5}{5}+\frac{2}{3}{n}^3-\frac{n}{105}$ is a natural number
So, we will check for $P(n+1)$
Consider, $\frac{(n+1{)}^7}{7}+\frac{(n+1{)}^5}{5}+\frac{2}{3}(n+1{)}^3-\frac{n+1}{105}$
$=\frac{n^7+7n^6+21n^5+35n^4+35n^3+21n^2+7n+1}{7}+\frac{n^5+5n^4+10n^3+10n^2+5n+1}{5}+\frac{2}{3}(n^3+3n^2+n+1)-\frac{n+1}{105}$
$=(\frac{(n+1{)}^7}{7}+\frac{(n+1{)}^5}{5}+\frac{2}{3}(n+1{)}^3-\frac{n+1}{105})+(n^6+3n^5+6n^4+7n^3+7n^2+4n)+(\frac{1}{7}+\frac{1}{5}+\frac{2}{3}-\frac{1}{105})$
$=P\left(n\right)+\text{natural number}+P\left(1\right)$
So, $P(n+1)$ is a natural number.
Hence, by mathematical induction method , the given result is true for all $n\in I^{+}$