To solve the equation:

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

let's simplify and solve for \( n \):

### Step 1: Find a common denominator

The least common denominator of the fractions on the left side is \( n^2 \). Rewrite each term with this common denominator:

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

So, the equation becomes:

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

### Step 2: Eliminate the fractions by multiplying by \( 2n^2 \)

Multiply both sides of the equation by \( 2n^2 \) to get rid of the denominators:

\[
2(1 + n) = 1
\]

Simplify the left side:

\[
2 + 2n = 1
\]

### Step 3: Solve for \( n \)

Subtract 2 from both sides:

\[
2n = 1 - 2
\]

\[
2n = -1
\]

Now, divide by 2:

\[
n = -\frac{1}{2}
\]

So, the solution is:

\[
n = -\frac{1}{2}
\]

Would you like a more detailed explanation, or do you have any questions?

### Related Questions:

1. What happens if \( n \) is substituted back into the original equation?
2. How would the solution change if the equation was \(\frac{1}{n^2} + \frac{1}{n} = \frac{1}{n^3}\)?
3. Can this equation be solved graphically, and what would the intersection points represent?
4. What is the behavior of the function \(\frac{1}{n^2} + \frac{1}{n} - \frac{1}{2n^2}\) as \( n \) approaches infinity?
5. How does the solution change if you change the fractions to decimals?

**Tip:** Always simplify equations by finding a common denominator to make calculations easier!

Explore how to solve the equation 1/(n²) + 1/n = 1/(2n²) step-by-step with detailed explanations. Learn about equations involving fractions and their solutions. Suitable for high school students.

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