# Show  1.2.3 + 2.3.4+ 3.4.5 + ... + n(n+1)(n+2) = 1/4n(n+1)(n+2)(n+3)?

Apr 1, 2017

The sum is:

${S}_{n} = 1.2 .3 + 2.3 .4 + 3.4 .5 + \ldots + n \left(n + 1\right) \left(n + 2\right)$
$\text{ } = {\sum}_{r = 1}^{n} r \left(r + 1\right) \left(r + 2\right)$
$\text{ } = {\sum}_{r = 1}^{n} r \left({r}^{2} + 3 r + 2\right)$
$\text{ } = {\sum}_{r = 1}^{n} \left({r}^{3} + 3 {r}^{2} + 2 r\right)$
$\text{ } = {\sum}_{r = 1}^{n} {r}^{3} + 3 {\sum}_{r = 1}^{n} {r}^{2} + 2 {\sum}_{r = 1}^{n} r$

We now use some standard formula for summation:

${\sum}_{r = 1}^{n} r \setminus = \frac{1}{2} n \left(n + 1\right)$
${\sum}_{r = 1}^{n} {r}^{2} = \frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right)$
${\sum}_{r = 1}^{n} {r}^{3} = \frac{1}{4} {n}^{2} {\left(n + 1\right)}^{2}$

Which gives us:

${S}_{n} = \frac{1}{4} {n}^{2} {\left(n + 1\right)}^{2} + 3 \cdot \frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right) + 2 \cdot \frac{1}{2} n \left(n + 1\right)$
$\text{ } = \frac{1}{4} {n}^{2} {\left(n + 1\right)}^{2} + \frac{1}{2} n \left(n + 1\right) \left(2 n + 1\right) + n \left(n + 1\right)$
$\text{ } = \frac{1}{4} n \left(n + 1\right) \left\{n \left(n + 1\right) + 2 \left(2 n + 1\right) + 4\right\}$>
$\text{ } = \frac{1}{4} n \left(n + 1\right) \left({n}^{2} + n + 4 n + 2 + 4\right)$
$\text{ } = \frac{1}{4} n \left(n + 1\right) \left({n}^{2} + 5 n + 6\right)$
$\text{ } = \frac{1}{4} n \left(n + 1\right) \left(n + 2\right) \left(n + 3\right) \setminus \setminus$ QED

Apr 1, 2017

We can also prove the given result using Mathematical Induction.

Let:

${S}_{n} = 1.2 .3 + 2.3 .4 + 3.4 .5 + \ldots + n \left(n + 1\right) \left(n + 2\right)$
$\text{ } = {\sum}_{r = 1}^{n} r \left(r + 1\right) \left(r + 2\right)$

We want to prove that:

${S}_{n} = \frac{1}{4} n \left(n + 1\right) \left(n + 2\right) \left(n + 3\right) \setminus \setminus \setminus \forall n \in \mathbb{N}$

Let use consider the case $n = 1$. When $n = 1$ the given result gives:

$L H S = {S}_{1}$
$\text{ } = 1.2 .3$
$\text{ } = 6$

And:

$R H S = \frac{1}{4} \left(1\right) \left(1 + 1\right) \left(1 + 2\right) \left(1 + 3\right)$
$\text{ } = \frac{1.2 .3 .4}{4}$
$\text{ } = \frac{24}{4}$
$\text{ } = 6$

So the given result is true when $n = 1$

Now, Let us assume that the given result is true when $n = k$, for some $k \in \mathbb{N}$, in which case for this particular value of $k$ we have:

${S}_{k} = {\sum}_{r = 1}^{k} r \left(r + 1\right) \left(r + 2\right) = \frac{1}{4} k \left(k + 1\right) \left(k + 2\right) \left(k + 3\right)$

Adding the next term to the series gives us:

${S}_{k + 1} = {S}_{k} + \left(k + 1\right) \left(k + 2\right) \left(k + 3\right)$
$\text{ } = \frac{1}{4} k \left(k + 1\right) \left(k + 2\right) \left(k + 3\right) + \left(k + 1\right) \left(k + 2\right) \left(k + 3\right)$ (by assumption)
$\text{ } = \frac{k \left(k + 1\right) \left(k + 2\right) \left(k + 3\right)}{4} + \frac{4 \left(k + 1\right) \left(k + 2\right) \left(k + 3\right)}{4}$
$\text{ } = \frac{1}{4} \left(k + 1\right) \left(k + 2\right) \left(k + 3\right) \left(k + 4\right)$
$\text{ } = \frac{1}{4} \setminus \left(k + 1\right) \setminus \left\{\left(k + 1\right) + 1\right\} \setminus \left\{\left(k + 1\right) + 2\right\} \setminus \left\{\left(k + 1\right) + 3\right\}$

Which is the given result with $n = k + 1$

So, we have shown that if the given result is true for $n = k$, then it is also true for $n = k + 1$. But we initially showed that the given result was true for $n = 1$ and so it must also be true for $n = 2 , n = 3 , n = 4 , \ldots$ and so on.

Hence, by the process of mathematical induction the given result is true for $n \in \mathbb{N}$ QED