2

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

I can get to $1/3(k+1)(k+2) + (k+1)(k+2)$ but then finishing off and rearranging the problem to $1/3(k+2)(k+3)$ is where I always get stuck.

Daniel W. Farlow
  • 22,531
user2733843
  • 61
  • 1
  • 6
  • 2
    I'm not sure the conclusion is correct. Is there a missing multiplier? A quick check goes like this: $n(n+1)=2\binom {n+1}2\to 2\sum_{i=1}^n\binom {i+1}2=2\binom{i+1}3$... – abiessu Jul 08 '16 at 20:32
  • This post on how to write a clear induction proof will likely serve you well. I tried to outline in an answer there one way of writing up your proofs that will force you to understand what you are doing. I suggest you give that a try. – Daniel W. Farlow Jul 08 '16 at 20:51
  • Did you try using Pascal's triangle? I'll give you a hint, start at the row that goes $1,7,\dots$, and go to the number $35$. Remember, it is the sum of the two numbers above it, $15+20$. Take the $20$, and remake it into the sum of the two numbers above it, $15+10+10$. Take the next two numbers above the $10$ on the right side, etc. We end up with $$35=1+3+6+10+15=35$$Multiply both sides by $2$ and we get$$70=1\cdot2+2\cdot3+3\cdot4+4\cdot5+5\cdot6$$Can you derive the formula from this? – Simply Beautiful Art Dec 09 '16 at 21:52

5 Answers5

2

$$\sum\limits_{i=1}^{n}{i(i+1)=}\sum\limits_{i=1}^{n}{{{i}^{2}}+\sum\limits_{i=1}^{n}{i=}}\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}=\frac{n(n+1)(n+2)}{3}$$

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

let

$$1(2)+2(3)+3(4)+...+k(k+1)=\frac 13 k(k+1)(k+2)$$ we show

$$1(2)+2(3)+3(4)+...+(k+1)(k+2)=\frac 13 (k+1)(k+2)(k+3)$$ Proof $$1(2)+2(3)+3(4)+...+k(k+1)+\color{red}{(k+1)(k+2)}=\frac 13 k(k+1)(k+2)+\color{red}{(k+1)(k+2)}$$ $$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad=\frac13 (k+1)(k+2)(k+3)$$

Behrouz Maleki
  • 11,182
2

Here is an easy way of going about the inductive step: \begin{align} \sum_{i=1}^{k+1}i(i+1)&= \sum_{i=1}^ki(i+1)+(k+1)(k+2)\\[1em] &= \frac{k(k+1)(k+2)}{3}+(k+1)(k+2)\\[1em] &= \frac{k(k+1)(k+2)+3(k+1)(k+2)}{3}\\[1em] &= \frac{(k+1)(k+2)(k+3)}{3}. \end{align} See if you can determine how each line was obtained (where the inductive hypothesis was applied, etc.).

Daniel W. Farlow
  • 22,531
  • Thank you for this, it is really helpful to see it written out like this but are you then saying that my textbook must be wrong because your answer is correct? It seems like you are probably right. – user2733843 Jul 08 '16 at 20:44
  • What makes you think your textbook and my answer oppose each other? Do you fully understand what you are actually trying to show using an inductive argument? For some statement $S(n)$, you first show that $S(1)$ is true (the base case). Then you assume that $S(k)$ is true for some fixed $k\geq1$ (the inductive hypothesis). Then you aim to show that $S(k)$ implies $S(k+1)$. My answer shows the main part, namely how $S(k)$ implies $S(k+1)$. – Daniel W. Farlow Jul 08 '16 at 20:49
1

When $n=1$, it is trival.

Suppose it is true for $n=k$, ie. $\sum \limits_i^k i(i+1)=\frac{k(k+1)(k+2)}{3}$.

Consider $n=k+1$. $$\begin{align} \sum \limits_i^{k+1} i(i+1)=&(k+1)(k+2)+\sum \limits_i^{k} i(i+1) \\ =&(k+1)(k+2)+\frac{k(k+1)(k+2)}{3} \\ =&\frac{(k^2+4k+3)(k+2)}{3} \\ =&\frac{(k+1)(k+1+1)(k+1+2)}{3}. \end{align} $$ Done!

Xianjin Yang
  • 677
  • Thank you for this, it is really helpful to see it written out like this but are you then saying that my textbook must be wrong because your answer is correct? It seems like you are probably right. (sorry for the copy and paste as above but you both answered very clearly). – user2733843 Jul 08 '16 at 20:45
  • What is the proof in your textbook? – Xianjin Yang Jul 08 '16 at 20:50
1

Your proposition is not true.

try $n = 2$

$1\cdot 2 + 2\cdot3 \ne \frac 13 3\cdot4$

Which is why you are having a tough time proving this by induction.

Induction is usefull to prove things, but it doesn't always have a whole lot of insight why things are the way they are.

This is probably what you should have:

$\sum_\limits{k=1}^{n} (k)(k+1) = \frac 13 n(n+1)(n+2)$

base case:

$n = 1\\ 1\cdot2 = \frac13 1\cdot2\cdot 3$

Inductive hypothesis:

$\sum_\limits{k=1}^{n} (k)(k+1) = \frac 13 n(n+1)(n+2)$

Show that

$\sum_\limits{k=1}^{n+1} (k)(k+1) = \frac 13 (n+1)(n+2)(n+3)$

based on the inductive hypothesis.

$\sum_\limits{k=1}^{n+1} (k)(k+1)\\ \sum_\limits{k=1}^{n} (k)(k+1) + (n+1)(n+2)\\ \frac 13 n(n+1)(n+2) + (n+1)(n+2)\\ (n+1)(n+2) (\frac 13 n + 1)\\ \frac 13(n+1)(n+2)(n + 3)$

QED

Doug M
  • 57,877
0

Let $ 2 = \frac{1 \cdot 2 \cdot 3}{3} $ be the basis. The inductive step consists of simple distribution of multiplication: $$ \begin{align}&\frac{n(n+1)(n+2)}{3} + (n+1)(n+2) \\&= \tfrac{1}{3}n^3+2n^2+\tfrac{11}{3}n+2 \\&= \frac{(n + 1)\Big((n+1)+1\Big)\Big((n + 1)+2\Big)}{3}\end{align} $$