2

I am looking for an induction proof...

$$2 + 5 + 8 + 11 + \cdots + (9n - 1) = \frac{3n(9n + 1)}{2}$$

when $n \geq 1$.

I know there are $3$ steps to this.
1) Check
2) Do $n = k$
3) Do $n = k + 1$

Problem is, I can't seem to get past the first step, when I check: $n = 1: (9(1) - 1) = 8$ but isn't it suppose to be come to $2$?

Michael Hardy
  • 1
Fenris
  • 37
  • $2+5+8 = 15 = \dfrac{3 \cdot 10}{2} = 15$. – dezdichado May 06 '16 at 01:06
  • 2
    Look carefully at the statement you are trying to prove. The first case is when $n=1$, which corresponds to $2+5+8=[31(91+1)]/2$. Note that the left hand side has 3 terms because it goes up to 9*1-1=8 by definition. – alphacapture May 06 '16 at 01:06
  • Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig May 06 '16 at 01:26
  • Each step adds three terms. n = 1 => 2 + 5 + 8. n = 2 => (2 + 5 + 8) + (11 + 14 + 17). n = n => (2+5 + 8) + ......+ [(9n - 7) + (9n - 4) + (9n -1)]. So n=1 => 2 + 5 + 8 = 3(9+1)/2 (which it does) and n=2 => 2 + 5 + 8 + 11 + 14 + 17 = 3x2(9x2 + 1)/2 which... 57 = 6(18 + 1)/2 = 3*19 = 57. Yep... it works. – fleablood May 06 '16 at 01:41
  • 1
    Maybe it would have been better written as (2 + 5 + 8) + (11 + 14 + 15) +....+ ((9n-7) + (9n -4) + (9n - 1)). – fleablood May 06 '16 at 01:43
  • So the induction step will be to show $\frac{3k(9k+1)}2 + (9k+2) + (9k+5) + (9k+8) = \frac{3(k+1)(9(k+1) + 1)}2$. – fleablood May 06 '16 at 01:48
  • fleablood I have a problem with your form: What if the parentheses don't match up when you get to the end? – alphacapture May 06 '16 at 10:52

4 Answers4

0

Let the sum be

$$S = 2 + 5 + 8 + 11 + ... + (9n - 4) + (9n - 1)$$

Let's add $S$ to itself

$$2S = [2 + (9n-1)] + [5 + (9n-4)] + ... + [(9n - 4) + 5] + [(9n - 1) + 2]$$

$$2S = [(9n+1)] + [(9n+1)] + ... + [(9n+1)] + [(9n + 1)]$$

So

$$2S = \left(\frac{(9n-1) - 2}{3} + 1\right)(9n+1)$$

$$S = \frac{3n(9n+1)}{2}$$

user1543042
  • 486
0

For $n=1$, Ok. It's easy to check

Now assume it works for $n$, the we have to show it works for $n+1$. (note that $9(n+1)-1=9n+8$)

we have to show the sentence below is true:

$$2+5+8+...+9n-1+9n+2+9n+5+9n+8=\frac{3(n+1)(9(n+1)+1)}{2}=\frac{27n^2+27n+30}{2}$$

Knowing it works for $n$, then we have:

$$2+5+8+...+9n-1+9n+2+9n+5+9n+8$$ $$=\frac{3n(9n+1)}{2}+9n+2+9n+5+9n+8$$ $$=\frac{27n^2+3n}{2}+27n+15$$ $$=\frac{27n^2+57n+30}{2}$$ $$=\frac{3(n+1)(9(n+1)+1)}{2}.$$

Curious
  • 369
0

The correct form of the sum is :

$\sum_{k=1}^{n}(2+3(k-1))=2n+3\sum_{k=0}^{n-1}k=2n+3\dfrac{(n-1)n}{2}$

$=\dfrac{3n^{2}+n}{2}$. The "$9n-1$" form is incorrect!!

(I mean in the question and hence in the answers)!

0

For $n=1$, since $9n-1=8$, on the the left hand side, you have $$2+5+8=15.$$ On the other hand, on the right hand side you have $$\frac{3n(9n+1)}{2}=\frac{3(10)}{2}=15$$ again. So the base of your induction $n=1$ is okay.

Bob Dobbs
  • 10,988