1. ## Logic Proof

hey does anyone know how to solve this?

Prove 1x2+2x3+3x4+...+n(n+1)=1/3n(n+1)(n+2) for each n of the element of natural numbers. x = multiplication

thanks...

2. Hello,

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

Do you know the general formulae for these sums ? You should have learnt it...i think ^^

3. Hello, TheMikego!

Prove: . $1\cdot2+2\cdot3+3\cdot4+ \hdots + n(n+1)\:=\:\frac{n(n+1)(n+2)}{3}\;\;\text{ for all }n \in N$

We have: . $S(n) \:=\:1\cdot2+2\cdot3+ 3\cdot4 + 4\cdot5 + \hdots + n(n+1)\;=\;\frac{n(n+1)(n+2)}{3}$

Verify $S(1)\!:\;\;1\cdot 2 \:=\:\frac{1\cdot2\cdot3}{3} \:=\:2\quad\hdots\quad\text{True!}$

Assume $S(k)\!:\;\;1\cdot2+2\cdot3+3\cdot4 + \hdots + k(k+1) \;=\;\frac{k(k+1)(k+2)}{3}$

Add $(k+1)(k+2)$ to both sides:

. . $\underbrace{1\cdot2+2\cdot3+3\cdot4+\hdots+(k+1)(k +2)}_{\text{This is the left side of }S(k+1)} \;=\;\frac{k(k+1)(k+2)}{3} + (k+1)(k+2)$

The right side is: . $\frac{k(k+1)(k+2)}{3} +\frac{{\color{blue}3}(k+1)(k+2)}{{\color{blue}3}}$

. . $\text{Factor: }\;\frac{(k+1)(k+2)}{3}\,[k+3] \;=\;\underbrace{\frac{(k+1)(k+2)(k+3)}{3}}_{\text {Right side of }S(k+1)}$

We have established $S(k+1)$ . . . The inductive proof is complete.

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# 1x2 2x3... n(n 1)=1/3n(n 1)(n 2)

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