## sasogeek 3 years ago prove by induction that $$\large 1 \times 2 + 2 \times 3 + ... +n(n+1) = \frac{1}{3}n(n+1)(n+2)$$

1. Mimi_x3

well, where are you stuck i assume you should know the steps n=k, then n=k+1

2. Mimi_x3

Assume $$n=k$$ is true $1*2+2*3+...+k(k+1) = \frac{1k}{3}(k+1)(k+2)$ Prove $$n=k+1$$ $1*2+2*3+...+k(k+1)+(k+1)(k+2) = \frac{(k+1)(k+2)(k+3)}{3}$ $\frac{(k)(k+1)(k+2)}{3} + (k+1)(k+2) = \frac{(k+1)(k+2)(k+3)}{3}$ Now, you can prove RHS = LHS

3. sasogeek

ok i understand and have worked the up to the point $$\large 1* 2 + 1* 3 + ... + k+1(k+2) = 1 * 2 + 1*3 + ... +k(k+1) +k+1(k+2)$$ $$\large = \frac{1}{3}k(k+1)(k+2)+(k+1)(k+2)$$ then what?

for left side : k(k+1)(k+2)/3 + (k+1)(k+2) = (k+1)(k+2)(k/3 + 1) = (k+1)(k+2)(k+3)/3 same like right side

5. Mimi_x3

Well, you prove as I said above. Prove the LHS = RHS $=>(k+1)(k+2)\left[ \frac{k}{3}+1\right]$It should be straight forward now.

6. Mimi_x3

Just, some algebra and you're done!

7. sasogeek

it's still blurry but i'll try to get it in a bit :)

8. Mimi_x3

Well, where are you stuck?

9. Mimi_x3

All you have to do here is: Prove the LHS that is: $\frac{(k)(k+1)(k+2)}{3} + (k+1)(k+2)$ Is equal to the RHS: $=> \frac{(k+1)(k+2)(k+3)}{3}$

10. sasogeek

ohh, thanks :) normal algebra takes off from there i see :) makes sense now xD