Hard math problem for CC peeps =)

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1

<p>do it up...unless you ain't man enough =P</p>

<p>Suppose you have a sequence where A(1) = 1, A(2) = 1, A(3) = 1, and then A(n+3) = A(n) + A(n+1) + A(n+2)...</p>

<p>now you're given that A(28) = 6090307, A(29) = 11201821, and A(30) = 20603361...</p>

<p>without using a calculator, and without brute forcing the entire series, find the sum of the first 30 terms of the series...</p>

<p>(keep in mind...figuring out how to work out the solution is more important than getting the correct answer at the end)</p>

2

<p>I don’t know if this is correct at all, but it’s what I have:</p>

<p>A(30)= 20603361
A(29)=11201821</p>

<p>A(30) - A(29)=9401540</p>

<p>A(29)=11201821
A(28)=6090307</p>

<p>A(29) - A(28)=5111514</p>

<p>So, I took 5111514 + 9401540</p>

<p>= 14513054</p>

<p>Then, A(30) - 14513054</p>

<p>= A(28)</p>

<p>I know I didn’t answer your question, but I’m pretty sure I’ve find the necessary missing numbers from the given pattern. Am I on the right track, at least?</p>

3

<p>Just to clarify, A(4)=A(1+3)=A(1)+A(2)+A(3) or A(1)+A(2)+A(3)+A(4)?</p>

4

<p>Missing numbers:</p>

<p>Going backwards:
Difference 1: 9401540
Difference 2: 5111514</p>

<p>D1 + D2 = 14513054 (clearly, the number has something to do with the problem)</p>

<p>D1 - D2 = 4290026 (could this be A(27)?)</p>

5

<p>I’m going to bed, but the way I’d do this is…</p>

<p>A4=a1+a2+a3
A5=a2+a3+a4 = a2+a3+a1+a2+a3 = a1+2a2+2a3
A6=a3+a4+a5 = a3+ a1+a2+a3+a2+a3+a1+a2+a3 = 2a1 + 3a2 + 4a3
A7=a4+a5+a6 = 4a1 + 6a2 + 7 A3
A8 = 7a1 + 11a2 + 13a3
.
.
.</p>

<p>It’s a sequence yay! I don’t know this seems like brute forcing it to me, and the numbers must get pretty big at the end, so I think I’m wrong. I thought A(27) was the key too at first but after three minutes of thinking I couldn’t figure out anything :(. It probably is though.</p>

6

<p>^ Easier than the crap I’m attempting to do. lol</p>

7

<p>

</p>

<p>I think by A(n+3) = A(n) + A(n+1) + A(n+2)…, OP means that A(27)=A(24+3)=A(24)+A(25)+A(26)+…A(24+24). I think. I mean, if a(n+3)=a(n)+a(n+1)+a(n+2), this problem wouldn’t be that hard. But if that were true, then there wouldn’t be ellipses after A(n+2) in the OP.</p>

8

<p>@jefgreen:

</p>

<p>@314159265, I did it another way it means A(4) = A(3) + A(2) + A(1).</p>

<p>Another method you could do is:
20603361/11201821 = 1.8392886755
11201821/6090307 = 1.8392885755
6090307*1.83928655 = 3311233</p>

<p>Essentially the sequence means 1.8392886755 * x 30 times well in this case 28 because A(1) and A(2) don’t count.</p>

<p>Answer = 1.8392885755 ^ (28) = 23338720.66</p>

<p>It is an approximate answer since at the beginning the percent change varies. I know it said not to use a calculator, but I couldn’t help myself. ;)</p>

<p>Hopefully someone else will solve it the correct way. :)</p>

9

<p>1 1
2 1
3 1
4 3
5 5
6 9
7 17
8 31
9 57
10 105
11 193
12 355
13 653
14 1201
15 2209
16 4063
17 7473
18 13745
19 25281
20 46499
21 85525
22 157305
23 289329
24 532159
25 978793
26 1800281
27 3311233
28 6090307
29 11201821
30 20603361</p>

10

<p>Solution:</p>

<p>Set up the following notation: S(m,n) = A(m) + A(m+1) + … + A(n)</p>

<p>A(1) + A(2) + A(3)

  • A(2) + A(3) + A(4)
  • A(3) + A(4) + A(5)
  • A(4) + A(5) + A(6)
  • …
    +A(28) + A(29) + A(30)</p>

<p>= A(4)

  • A(5)
  • A(6)
  • A(7)
  • …
  • A(31)</p>

<p>Simplifying both sides using our notation, we have the equality:
3S(3,28) + A(1) + 2A(2) + 2A(29) + A(30) = S(4,31)</p>

<p>Note: S(4,31) = S(3,28) + A(29) + A(30) + A(31) - A(3)</p>

<p>Thus, 3S(3,28) + A(1) + 2A(2) + 2A(29) + A(30) = S(3,28) + A(29) + A(30) + A(31) - A(3)</p>

<p>So S(3,28) = (A(31) - A(29) - 2A(2) - A(1) - A(3))/2</p>

<p>Note: S(1,30) = A(1) + A(2) + S(3,28) + A(29) + A(30)</p>

<p>So S(1,30) = (A(31) + A(29) + A(1) - A(3) + 2A(30))/2</p>

<p>= (A(31) + A(29) + 2A(30))/2</p>

<p>= (A(28) + A(29) + A(30) + A(29) + 2A(30))/2</p>

<p>= (A(28) + 2A(29) + 3A(30))/2</p>

<p>= (6090307 + 2<em>11201821 + 3</em>20603361)/2 </p>

<p>= 45152016</p>

11

<p>A more clearly motivated solution would be (same notation):</p>

<p>S(1,30) = A(1) + A(2) + A(3) + A(4) + A(5) + A(6) + … + A(29) + A(30)</p>

<p>= A(1) + A(2) + A(3) + (A(1) + A(2) + A(3)) + (A(2) + A(3) + A(4)) + (A(3) + A(4) + A(5)) + … + (A(26) + A(27) + A(28)) + (A(27) + A(28) + A(29))</p>

<p>= 2A(1) + 2A(2) + 3A(3) + 3A(4) + 3A(5) + … + 3A(27) + 2A(28) + A(29)</p>

<p>So S(1,30) = 2A(1) + 2A(2) + 3S(3,27) + 2A(28) + A(29)</p>

<p>Note: S(3,27) = S(1,30) - A(1) - A(2) - A(28) - A(29) - A(30)</p>

<p>So S(1,30) = 2A(1) + 2A(2) + 3(S(1,30) - A(1) - A(2) - A(28) - A(29) - A(30)) + 2A(28) + A(29)</p>

<p>So S(1,30) = (A(28) + 2A(29) + 3A(30))/2</p>

<p>= (6090307 + 2<em>11201821 + 3</em>20603361)/2 </p>

<p>= 45152016</p>

12

<p>damn you guys really went all out with this…i’m wayyy too lazy to read all your solutions…</p>

<p>part of it is realizing that A(30) can be expanded to be the sum of all the terms that aren’t the multiple of 3
A(30) = A(29) + A(28) + A(27)
A(27) = A(26) + A(25) + A(24)</p>

<p>therefore</p>

<p>A(30) = A(29) + A(28) + A(26) + A(25) + A(24)</p>

<p>then A(24) could be re-written…this pattern follows…you can then write A(29) and A(28) in similar ways…then you just play around with it (i think it was…if you add the long hand expression of all three, you get something like 2x the sum of terms 1 - 28, and then term 29…if you add in term 29 again, and term 30 two more times, you get double the sum of the whole thing)…</p>

13

<p>yeah Simo14 nailed it…i haven’t looked at the other solutions lol</p>