Question

prove by induction that
\(\large 1 \times 2 + 2 \times 3 + ... +n(n+1) = \frac{1}{3}n(n+1)(n+2) \)

Answer 1

well, where are you stuck

i assume you should know the steps n=k, then n=k+1

Answer 2

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

Answer 3

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?

Answer 4

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

Answer 5

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.

Answer 6

Just, some algebra and you're done!

Answer 7

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

Answer 8

Well, where are you stuck?

Answer 9

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:
\[=> \frac{(k+1)(k+2)(k+3)}{3}\]

Answer 10

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