Question

Use induction to prove that the following equality holds for all natural numbers n:
the summation from r=1 to n of r(r+5)= (n(n+1)(n+8))/3

Answer 1

I have proven the base case but I am having alot of trouble with the induction step.

Answer 2

have you ever done a proof by induction before?

Answer 3

yes, I'm just having trouble making it look like the right form during the induction step. I keep getting it incorrect

Answer 4

ok my guess is the algebra is going to suck

Answer 5

you want me to see if i can work it out?
if i try just typing it i i will probably make a mistake

Answer 6

the gist if it is going to be to show that
\[(k+1)(k+6)+\frac{k(k+1)(k+8)}{3}=\frac{(k+1)(k+2)(k+9)}{3}\]
did you get to that step?

Answer 7

I got stuck somewhere in there haha! i got the first part but getting fromt he first part to the second I must have made a mistake

Answer 8

ok well if you got there, and assuming the answer is right, the algebra cannot be that hard really, since we know what we are aiming for

Answer 9

\[(k+1)(k+6)+\frac{k(k+1)(k+8)}{3}\]
\[=\frac{3(k+1)(k+6)+k(k+1)(k+6)}{3}\] is a start

Answer 10

also a typo,should be \[=\frac{3(k+1)(k+6)+k(k+1)(k+8)}{3}\]

Answer 11

ok I will keep working then, now that i have an answer to work towards. Thanks!

Answer 12

you can multiply all that mess out, but it might be easier to factor out the common factor of \(k+1\) in the numerator, because you have that factor on the right hand side as well

Answer 13

you would get, working this way
\[(k+1)[3(k+6)+k(k+8)]\] just up top
lets see if we can make it work fast

Answer 14

oh heck yes, works nicely this way
\[(k+1)(3(k+6)+k(k+8)\]
\[(k+1)(3k+18+k^2+8k)\]
\[=(k+1)(k^2+11k+18)=(k+1)(k+2)(k+9)\] as needed
not that bad after all!

Answer 15

you are so awesome! thank you!

Answer 16

lol it is just boring algebra, but if you did not use the common factor of \(k+1\) in the top the algebra would stink as you would have to work with cubes
so i will take credit for that
yw