Decompose 1/(2n^2 + n) into partial fractions.

Step-by-step solution

1. 1

   Factor the denominator. Both terms share $n$:

   $2n^2+n=n(2n+1)\quad\Longrightarrow\quad\frac{1}{2n^2+n}=\frac{1}{n(2n+1)}$

   Two distinct linear factors means one term each, with no repeated powers.

2. 2

   Set up the template and clear denominators.

   $\frac{1}{n(2n+1)}=\frac{A}{n}+\frac{B}{2n+1}\quad\Longrightarrow\quad 1=A(2n+1)+Bn$

3. 3

   Group by powers of $n$ and equate coefficients.

   $1=(2A+B)n+A$

   The constant term gives $A=1$; the coefficient of $n$ must vanish, so $2A+B=0$ and therefore

   $B=-2A=-2$

4. 4

   Write the decomposition.

   $\frac{1}{2n^2+n}=\frac{1}{n}-\frac{2}{2n+1},\qquad n\neq 0,\ n\neq-\tfrac12$

5. 5

   Check by recombining and numerically. $\tfrac{(2n+1)-2n}{n(2n+1)}=\tfrac{1}{n(2n+1)}$ $\checkmark$. At $n=1$: left $=\tfrac13$, right $=1-\tfrac23=\tfrac13$ $\checkmark$. At $n=2$: left $=\tfrac{1}{10}$, right $=\tfrac12-\tfrac25=\tfrac{1}{10}$ $\checkmark$. This decomposition is the standard first step for summing the series $\sum\tfrac{1}{2n^2+n}$.