use induction to prove that $\sum_{k=1}^{n} k(k + 1) = \frac{n(n + 1)(n + 2)}{3}$ for all $n ∈ N$.

Question

I am supposed to use induction to prove that

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

for all $n \in \mathbb{N}$.

I am confused on how to go about this question, a step by step guide of this proof will be helpful.

Answer 1

For $n=1$ you get $2=2$

If true for n, then we know $$ \sum_{k=1}^{n} k(k + 1) = [n(n + 1)(n + 2)]/ 3$$

Thus $$ \sum_{k=1}^{n+1} k(k + 1) = [n(n + 1)(n + 2)]/ 3 +(n+1)(n+2) $$

$$ = [n(n + 1)(n + 2) +3(n+1)(n+2)]/3 = [(n + 1)(n + 2)(n+3)]/ 3$$

Which is the statement for $n+1$

Answer 2

That is a well known property of Rising Factorials.
To prove it by induction, we follow the standard steps.

a) the thesis is true for $n=1$

$$ \sum\limits_{k = 1}^1 {k\left( {k + 1} \right)} = 2 = {{1 \cdot 2 \cdot 3} \over 3}\quad :\quad TRUE $$

b) if it is true for $n$ it is true for $n+1$ $$ \eqalign{ & \sum\limits_{k = 1}^{n + 1} {k\left( {k + 1} \right)} = \sum\limits_{k = 1}^n {k\left( {k + 1} \right)} + \left( {n + 1} \right)\left( {n + 2} \right) = \cr & = {{n\left( {n + 1} \right)\left( {n + 2} \right)} \over 3} + \left( {n + 1} \right)\left( {n + 2} \right) = \cr & = {{n\left( {n + 1} \right)\left( {n + 2} \right) + 3\left( {n + 1} \right)\left( {n + 2} \right)} \over 3} = \cr & = {{\left( {n + 3} \right)\left( {n + 1} \right)\left( {n + 2} \right)} \over 3} = \cr & = {{\left( {n + 1} \right)\left( {\left( {n + 1} \right) + 1} \right)\left( {\left( {n + 1} \right) + 2} \right)} \over 3}\quad :\quad TRUE \cr} $$

c) conclusion: the thesis is true for $1 \le n$

Answer 3

Let's try iteration:

$1\times 2=2=\frac{2\times 3}{3}$

$1\times 2+2\times3=8=\frac{2\times3\times 4}{3}$

$1\times 2+2\times3+3\times 4= 20=4\times 5=\frac{3\times 4\times 5}{3}$

$1\times 2+2\times3+3\times 4+4\times 5= 40=4\times 5\times 2=\frac{4\times 5\times 2\times 3}{3}=\frac{4\times 5\times 6}{3}$

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$1\times 2+2\times3+3\times 4+4\times 5 + . . .n(n+1)=\frac{n(n+1)(n+2)}{3}$