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

Options

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

Answer

To prove that the expression

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

is a natural number for all \( n \in \mathbb{N} \), we will use the principle of mathematical induction.

### Step 1: Base Case
We start by checking if the statement is true for \( n = 1 \).

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

Calculating each term:

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

To add these fractions, we need a common denominator. The least common multiple of 5, 3, and 15 is 15.

Now, we convert each term:

\[ \frac{1}{5} = \frac{3}{15}, \quad \frac{1}{3} = \frac{5}{15}, \quad \frac{7}{15} = \frac{7}{15} \]

Adding these fractions:

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

Since 1 is a natural number, the base case holds true.

### Step 2: Inductive Step
Assume that the statement is true for some \( n = k \), i.e.,

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

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

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

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

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 expansions into \( P(k + 1) \):

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

Now, we can separate the terms:

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

Grouping the terms:

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

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

### Step 3: Conclusion
The additional terms are all composed of \( k \) which is a natural number, and constants. Therefore, the sum of a natural number and a combination of natural numbers will also be a natural number.

Thus, by the principle of mathematical induction, \( P(n) \) is a natural number for all \( n \in \mathbb{N} \).

### Final Answer
The expression \( \frac{n^5}{5} + \frac{n^3}{3} + \frac{7n}{15} \) is a natural number for all \( n \in \mathbb{N} \).