`(sum_(r=1)^n r^4)/(sum_(r=1)^n r^2)` is equal to

Answer

To solve the problem \(\frac{\sum_{r=1}^n r^4}{\sum_{r=1}^n r^2}\), we need to find the formulas for both the summation of \(r^4\) and \(r^2\).

### Step 1: Find the formula for \(\sum_{r=1}^n r^2\)

The formula for the sum of the squares of the first \(n\) natural numbers is:

\[ \sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6} \]

### Step 2: Find the formula for \(\sum_{r=1}^n r^4\)

The formula for the sum of the fourth powers of the first \(n\) natural numbers is:

\[ \sum_{r=1}^n r^4 = \frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30} \]

### Step 3: Substitute the formulas into the expression

Now, we substitute the formulas into the expression:

\[ \frac{\sum_{r=1}^n r^4}{\sum_{r=1}^n r^2} = \frac{\frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30}}{\frac{n(n+1)(2n+1)}{6}} \]

### Step 4: Simplify the expression

To simplify this, we can cancel out the common terms in the numerator and denominator:

\[ = \frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30} \cdot \frac{6}{n(n+1)(2n+1)} \]

This simplifies to:

\[ = \frac{6(3n^2 + 3n - 1)}{30} \]

### Step 5: Further simplify the fraction

Now, we can simplify \(\frac{6}{30}\):

\[ = \frac{1}{5}(3n^2 + 3n - 1) \]

Thus, the final result is:

\[ = \frac{3n^2 + 3n - 1}{5} \]

### Final Answer

\[ \frac{\sum_{r=1}^n r^4}{\sum_{r=1}^n r^2} = \frac{3n^2 + 3n - 1}{5} \]

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