The formula provided in the image is:

$x_m = \frac{m(1 + 2 + \dots + m)}{1^2 + 2^2 + \dots + m^2}$

To simplify and understand this formula:

1. The numerator involves the sum of the first $m$ positive integers: $1 + 2 + \dots + m = \frac{m(m+1)}{2}$

2. The denominator involves the sum of the squares of the first $m$ positive integers: $1^2 + 2^2 + \dots + m^2 = \frac{m(m+1)(2m+1)}{6}$

By substituting these formulas into $x_m$:

$x_m = \frac{m \cdot \frac{m(m+1)}{2}}{\frac{m(m+1)(2m+1)}{6}}$

Simplify step by step:

• The numerator becomes: $\frac{m^2(m+1)}{2}$

• The denominator becomes: $\frac{m(m+1)(2m+1)}{6}$

Dividing the two expressions:

$x_m = \frac{\frac{m^2(m+1)}{2}}{\frac{m(m+1)(2m+1)}{6}}$

Simplify further:

$x_m = \frac{m^2(m+1) \cdot 6}{2 \cdot m(m+1)(2m+1)}$

Cancel common terms:

$x_m = \frac{3m}{2m+1}$

Final Simplified Formula:

$x_m = \frac{3m}{2m+1}$

Do you want details about the derivation steps, or do you have any questions?

Here are 5 related questions to explore further:

1. What is the significance of $x_m$ in this formula?
2. How can this formula be applied to specific problems in mathematics or physics?
3. What is the behavior of $x_m$ as $m \to \infty$?
4. Can you calculate $x_m$ for specific small values of $m$ (e.g., $m = 1, 2, 3$)?
5. How does the simplification process change if higher powers (e.g., cubes) replace squares in the denominator?

Tip:

Always substitute and simplify summation formulas before attempting to directly compute values. It makes the calculations more manageable!