Can someone help me figure out the pattern in this series of number?

[Hoeboy]

Find the sum of 7+13+19+...+673.

I know you're adding every 6th number and I can sit here for the next hour doing it manually but I don't think I can do that I really suck at patterns and deriving equations from patterns. Can someone give me a hint if you know how to do this??

7---13---19----25---31-------673
7---20---39----64---95-------???

 

[Orsorum]

1+6n, infinity (sigma) n=1.

 

[Hoeboy]

1+6n gives you the first line which basically says every 6th number. I need to know the sum up to 6n+1=673

 

[Orsorum]

n=1 to n=112.

 

[HonkeyDonk]

Why don't you use your calculator if you have one...like a Ti89 or 83

just do the Sigma notation of (1+6x) where x = 1 and goes to 112.

 

[Hoeboy]

The problem is this chapter is based on deriving equations from patterns. I am sure I won't get credit if I just plop the answer w/o any work.

 

[Orsorum]

Grrr....

 

[Orsorum]

Originally posted by: HonkeyDonk
Why don't you use your calculator if you have one...like a Ti89 or 83

just do the Sigma notation of (1+6x) where x = 1 and goes to 112.

Or you could take the integral of it...

 

[Hoeboy]

so basically it would look something like this


n---------1----2----3----4----5----6-----...-----112
6n+1----7----13--19--25---31---37---...-----673
Sum-----7----20--39--64---95---132---------????


i'm trying to find that ??? by deriving a formula in the pattern. the problem is i don't friggin see a pattern

 

[Orsorum]

x+3x^2?

112 + 37632?

 

[Basilisk6]

6n+1 from n=1 to n=112:

The equation for a sum from 1 to n is "n*(n+1)/2", so
112+6(112*113/2)

I think that's right, but you can check and solve it yourself.

Edit: The equation would be "n+6(n(n+1)/2)" where n=112

 

[Hoeboy]

wow good job basilisk. how the hell did you manage to see the pattern and derive that equation?

 

[Basilisk6]

I didn't actually see a pattern in the numbers themselves. I knew the sums equation(n(n+1)/2), and the numbers you're adding are:
(1+6(1))+(1+6(2))...(1+6(112)) (as Zakath15 pointed out)

You move the numbers around to get (1+1+...+1)+6(1+2+3+...+112), so (1+2+...+112) = (112(113)/2)

 

[Hoeboy]

im still a bit cloudy. and i'm supposed to be a math major. let me hang my head in shame

 

[Basilisk6]

Originally posted by: Hoeboy
im still a bit cloudy. and i'm supposed to be a math major. let me hang my head in shame

What are you cloudy about? I'll try to make it clearer.

 

[Hoeboy]

how did you get (n(n+1)/2)?

 

[Basilisk6]

Originally posted by: Hoeboy
how did you get (n(n+1)/2)?

It's a very well-known equation for the sum of the first n natural numbers. Here's a proof of it.

 

[hamburglar]

what's the answer?
I wrote a program but i wanna know if i did it right. Im not sure how many of you know scheme but here it is.

STk> (define (sum start add end)
(if (= start end)
start
(+ start (sum (+ add start) add end))))
sum
STk> (sum 7 6 673)
38080

 

[kherman]

If this is for a college level calculus course, you have to do the work. I rfor get how, but it's a summation. You have to find hte equivilent formula of the summantion. Been too long since I had to deal with this kind of math.

1. 7
2. 20
3. 39
4. 64

If you can find the formula for this, it woill probably work for.
1000. ????????

n+??
1. 1 + 6
2. 2 + 18
3. 3 + 36
4. 4 + 60

Looks alot eassier this way, doesn't it?
further:
1. 1 + 6*(1)
2. 2 + 6*(3)
3. 3 + 6*(6)
4. 4 + 6*(10)

From above
1 --> (1)
2 --> (3)
3 --> (6)
4 --> (10)

(n(n+1))/2

Not how it should be done in a calculaus course, but i did show the logic of how to solve it!
ANSWER:
n+6*[(n(n+1))/2]

EDIT:
When does n correspond to 673. Fianlly, an easy one.
1+6n=673
6n=672
n=672/6
n=102

so your asnwer is:
102+6*[(102(102+1))/2]
102+6*[(102*103)/2]
102+6*[(10506)/2]
102+6*5253
102+31518
31620

 

[ScrewFace]

You guys are browners! I thought I was good in math but you guys humbled me!