Proof by Induction:3+6+9+...+3n=3n(n+1)/2

[Linda]

Can anyone help me solve this problem:

3+6+9+...+3n = 3n (n+1)/2

Any help and explanation of how you solved would be appreciated. Thanks Linda

 

[stapel]

Can anyone help me solve this problem:

3+6+9+...+3n = 3n (n+1)/2

Any help and explanation of how you solved would be appreciated. Thanks Linda

What formulas have they given you? Also, your subject line mentions "proof by induction". Where are you stuck in that process? (here) For instance, your book probably showed how to prove that 1 + 2 + 3 + ... + n = (n (n + 1)) / 2. Modelling this, what have you done?

Please be complete. Thank you!

 

[Linda]

I am asked to subtract n from each side of the equation. I also must explain why this can be done on each term in the summation. Thank you

 

[HallsofIvy]

If I were doing this problem the first thing I would do is divide both sides by 3 to get 1+ 2+ 3+ ...+ n= n(n+1)/2 that staple mentions.


However, if you don't want to do that, induction always has two parts:

1) Show this is true for 1: with n= 1, "3+ 6+ 9+ ...+ 3n" becomes just "3". What do you get setting n= 1 in 3n(n+1)/2?

2) Assume the statement is true for some "k" and prove it is true for "k+1".

If 3+ 6+ 9+ ...+ 3k= 3k(k+ 1)/2 then
3+ 6+ 9+ ...+ 3k+ 3(k+ 1)= (3+ 6+ 9+ ...+ 3k)+ 3(k+1)= 3k(k+1)/2+ 3(k+ 1).

What do you get when you add 3k(k+1)/2+ 3(k+ 1)?

 

[stapel]

I am asked to subtract n from each side of the equation.

By whom? Why?

I also must explain why this can be done on each term in the summation.

:shock: I feel like I've walked in on a conversation, the first half of which I've missed.

Please reply with all of the information for this exercise, showing all of your work so far, so we can figure out what you're talking about. Thank you!