For all `n in N, (n^(5))/(5)+(n^(3))/(3)+(7)/(15n)` is

Options

A. an integer
B. a natural number
C. a positive fraction
D. none of these

Answer

To prove that for all \( n \in \mathbb{N} \), the expression

\[ P(n) = \frac{n^5}{5} + \frac{n^3}{3} + \frac{7}{15n} \]

is a natural number, we will use mathematical induction.

### Step-by-Step Solution:

**Step 1: Base Case**

We start by checking the base case, \( n = 1 \).

\[ P(1) = \frac{1^5}{5} + \frac{1^3}{3} + \frac{7}{15 \cdot 1} \]

Calculating each term:

\[ P(1) = \frac{1}{5} + \frac{1}{3} + \frac{7}{15} \]

To add these fractions, we find a common denominator, which is 15:

\[ P(1) = \frac{3}{15} + \frac{5}{15} + \frac{7}{15} = \frac{3 + 5 + 7}{15} = \frac{15}{15} = 1 \]

Since \( 1 \) is a natural number, the base case holds.

**Step 2: Inductive Hypothesis**

Assume that the statement is true for some \( n = k \), i.e.,

\[ P(k) = \frac{k^5}{5} + \frac{k^3}{3} + \frac{7}{15k} \text{ is a natural number.} \]

**Step 3: Inductive Step**

We need to show that \( P(k+1) \) is also a natural number.

Calculating \( P(k+1) \):

\[ P(k+1) = \frac{(k+1)^5}{5} + \frac{(k+1)^3}{3} + \frac{7}{15(k+1)} \]

Using the binomial theorem, we expand \( (k+1)^5 \) and \( (k+1)^3 \):

\[ (k+1)^5 = k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1 \]
\[ (k+1)^3 = k^3 + 3k^2 + 3k + 1 \]

Substituting these into \( P(k+1) \):

\[ P(k+1) = \frac{k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1}{5} + \frac{k^3 + 3k^2 + 3k + 1}{3} + \frac{7}{15(k+1)} \]

Now, we can separate the terms:

\[ P(k+1) = \left(\frac{k^5}{5} + \frac{10k^3}{5} + \frac{5k}{5} + \frac{1}{5}\right) + \left(\frac{k^3}{3} + \frac{3k^2}{3} + \frac{3k}{3} + \frac{1}{3}\right) + \frac{7}{15(k+1)} \]

Combining like terms:

\[ P(k+1) = P(k) + \left(\frac{5k^4 + 10k^2 + 1}{5}\right) + \left(\frac{3k^2 + 3k + 1}{3}\right) + \frac{7}{15(k+1)} \]

Since \( P(k) \) is a natural number by the inductive hypothesis, we need to show that the additional terms also yield a natural number.

**Step 4: Conclusion**

The terms \( \frac{5k^4 + 10k^2 + 1}{5} \) and \( \frac{3k^2 + 3k + 1}{3} \) are integers for all \( k \in \mathbb{N} \) because they are sums of integers divided by integers that evenly divide the sums.

Thus, \( P(k+1) \) is a natural number.

By the principle of mathematical induction, since the base case holds and the inductive step is valid, we conclude that

\[ P(n) = \frac{n^5}{5} + \frac{n^3}{3} + \frac{7}{15n} \]

is a natural number for all \( n \in \mathbb{N} \).