How do I prove that for $\forall n\in \mathbb{N}$ $\sum_{k=1}^{n}k(k+1)=\frac{1}{3}n(n+1)(n+2)$?

Question

How do I prove that for $\forall n\in \mathbb{N}$ $\sum_{k=1}^{n}k(k+1)=\frac{1}{3}n(n+1)(n+2)$?

I need to use induction.

1. For example if n=1. Than 2=2.

2. If statement holds for $\forall n\in \mathbb{N}$ then follows for $n + 1$

   $\sum_{k=1}^{n+1}k(k+1)=(n+1)(n+2)+\sum_{k=1}^{n}k(k+1)=\frac{1}{3}(n+1)(n+2)(n+3)$.

But what I prove with this?

Answer 1

You have proven that it is true for the case that n = 1 (i.e. that 2 = 2 - this is called the base case). You have then shown that if the relationship is true for n, it is also true for n + 1 (this is the inductive case); this is what the last line shows. Therefore, since it is true for n = 1, it is true for n + 1 = 2; since it is true for n = 2, it is true for n + 1 = 3, ... and hence it is true for all of the natural numbers.

Answer 2

You already have the first step, that it is true for $n=1$. In the second step, you are assuming it is true for $n$ and you want to prove it is true for $n+1$. So by assumption your $\sum_{k=1}^{n} k(k+1)$ is $\frac{1}{3} n (n+1)(n+2)$, and you want to show that

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

which is a straightforward calculation.

Answer 3

You have actually proved the statement, but it is a little hard to see, because in your inductive step you went from $n$ to $n+1$. Instead, go from $n-1$ to $n$:

Suppose the statement holds for $n-1$, so $$ S_{n-1} := \sum_{k=1}^{n-1} k (k+1) = {1 \over 3} (n-1) (n) (n+1). $$ Then $$ S_{n-1} + n(n+1) = n (n+1) \left[ 1 + {1 \over 3} (n-1) \right] = n (n+1) \left[ {n + 2 \over 3}\right]. $$

Answer 4

You have proved the statement via induction. Your job is done.

When you write that $$\sum_\limits{k}^{n+1}k(k+1)=\frac{1}{3}(n+1)(n+2)(n+3)$$ What you actually mean is $$\sum_\limits{k}^{n+1}k(k+1)=\frac{1}{3}(n+1)\{(n+1)+1\}\{(n+2)+1\}$$

So if I represent the mathematical statement $\sum_\limits{k}^{n}k(k+1)=\frac{1}{3}n(n+1)(n+2)$ as $P(n)$, then you have shown that if $P(n)$ is true, then $P(n+1)$ is also true.

And you have already proven that $P(1)$ is true.

Thus, from the above statement, we have
Truth of $P(1)$ $\implies$ Truth of $P(2)$
Truth of $P(2)$ $\implies$ Truth of $P(3)$
Truth of $P(3)$ $\implies$ Truth of $P(4)$
Truth of $P(4)$ $\implies$ Truth of $P(5)$
$\ldots$
$\ldots$
Truth of $P(n)$ $\implies$ Truth of $P(n+1)$ for all $n \in \mathbb{N}$

Hence the mathematical statement $$P(n):\sum_\limits{k}^{n}k(k+1)=\frac{1}{3}n(n+1)(n+2)$$ is true by the principle of mathematical induction.

Answer 5

Divide both sides by 2, and you have the fact that the sum up to $n+1$ of $ \left( \begin{array}{c} k \\ 2 \end{array} \right) $ is $ \left( \begin{array}{c} n+2 \\ 3 \end{array} \right) .$ This is the reason that the total number of gifts given in The Twelve Days of Christmas song is $ \left( \begin{array}{c} 14 \\ 3 \end{array} \right) =364 .$ In Pascal's triangle, this is summing along a diagonal: $$ 1 + 3 + 6 + 10 + \cdots + 66 + 78 = 364 $$