zarkam21
  • zarkam21
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Mathematics
katieb
  • katieb
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anonymous
  • anonymous
???????
anonymous
  • anonymous
with
zarkam21
  • zarkam21
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SolomonZelman
  • SolomonZelman
So, first you add 1. Then, you add 2. Then, add 4.
zarkam21
  • zarkam21
Okay so for number 4 the answer would be 25?
SolomonZelman
  • SolomonZelman
\(a_1=3\) \(a_2=4\) \(a_3=6\) \(a_4=10\) \(a_5=18\) \(a_6=34\) and so forth... \(a_2=a_1+1\) \(a_3=a_2+2=(a_1+1)+2=a_1+3=a_1+1+2\) \(a_4=a_3+4=(a_1+3)+4=a_1+7=a_1-1+2(4)=a_1-1+2^{4-1}\) \(a_5=a_4+8=(a_1+7)+8=a_1+15=a_1-1+2(8)=a_1-1+2^{5-1}\) \(a_6=a_5+16=(a_1+15)+16=a_1+31=a_1-1+2(16)=a_1-1+2^{6-1}\) \(a_n=a_1-1+2^{n-1}\)
SolomonZelman
  • SolomonZelman
I played around a little bit, and derived the following formula for the sequence, just in case you need it.
SolomonZelman
  • SolomonZelman
and you can further simplify this, knowing that \(a_1=3\) \(a_n=2+2^{n-1}\)
SolomonZelman
  • SolomonZelman
And the number of marbles in any \(n\)th line is given by \(a_n\).
zarkam21
  • zarkam21
So for the 7th line, there will be 50?
SolomonZelman
  • SolomonZelman
2+2\(^{7-1}\) 2+2\(^{6}\) 2+(2\(^{3}\))\(^2\) 2+8\(^2\) 2+64=66 this is a by-hand calculation.
zarkam21
  • zarkam21
okay so 66?
SolomonZelman
  • SolomonZelman
\(a_7\)=66
SolomonZelman
  • SolomonZelman
yes
zarkam21
  • zarkam21
Oh okay I understand the pattern now,, thanks ! :]
SolomonZelman
  • SolomonZelman
yw

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