Algebraic fractions: addition

Question

I have very elementary question about adding algebraic fractions. Now, I know the following:

$$\frac{a}{c} + \frac{b}{d} = \frac{da + cb}{dc}$$

My question is however, how given this expression:

$$\frac{(2n + 1) \; n(n + 1)}{2} + \frac{(2n + 1) \; n(n + 1)}{6}$$

One arrives at: $$\frac{(2n + 1) \; n(n + 1)}{3}$$

Trying to work out the solution, I arrive at a very large equation divided by 12...

Answer 1

$$\frac{(2n + 1) n(n + 1)}{2} + \frac{(2n + 1) n(n + 1)}{6} = (2n+1)n(n+1) \cdot \left( \frac{1}{2} + \frac{1}{6} \right) = \frac{2(2n+1)n(n+1)}{3}.$$

Answer 2

One should not arrive at that value, as it is not true. Notice that the numerators are the same. This is essentially

$$\frac{(2n+1)n(n+1)}{2} + \frac{(2n+1)n(n+1)}{6} = (2n+1)n(n+1) \left(\frac12 + \frac16\right).$$

Now, it is easy to show that $\frac12+\frac16 = \frac23$, so you should get

$$\frac{2(2n+1)n(n+1)}{3}.$$

Answer 3

$$\begin{align}\frac{(2n + 1) \; n(n + 1)}{2} + \frac{(2n + 1) \; n(n + 1)}{6} &= \frac{3(2n + 1) \; n(n + 1)}{6}+\frac{(2n + 1) \; n(n + 1)}{6} \\ &= \frac{4(2n + 1) \; n(n + 1)}{6} \\ &= \frac{2(2n + 1) \; n(n + 1)}{3} \end{align}$$

The equality is obtained by multiplying the first term by $1=\frac{3}{3}$.

The second is obtained by adding the numerators (as the denominators are the same).

The third is by noticing $\frac{4}{6}=\frac{2}{3}$

Answer 4

Hint: $ \dfrac{A}{2}+\dfrac{A}{6}=\dfrac{3A+A}{6}=\dfrac{4A}{6}=\dfrac{2A}{3} $ and your $A$ is ......