How can you find the upper bound of a geometric series like \(T = 7^1, 7^2, \ldots, 7^n\)?
Note 1: ETH::A&D
Deck: ETH::A&D
Note Type: Horvath Classic
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modified
Note Type: Horvath Classic
GUID:
m{e.+pS,Zn
Before
Front
Back
How can you find the upper bound of a geometric series like \(T = 7^1, 7^2, \ldots, 7^n\)?
Use the multiply-subract trick.
- Mutliply the series by its base: \(7T\)
- Subtract: \(7T - T = 7^{n+1} - 7^1\) (middle terms cancel)
- Factor: \(T(7-1) = 7^{n+1} - 7^1\)
- Divide: \(T = \frac{7^{n+1} - 7^1}{6}\)
After
Front
How can you find the upper bound of a geometric series like \(T = 7^1, 7^2, \ldots, 7^n\)?
Back
How can you find the upper bound of a geometric series like \(T = 7^1, 7^2, \ldots, 7^n\)?
Use the multiply-subract trick.
- Mutliply the series by its base: \(7T\)
- Subtract: \(7T - T = 7^{n+1} - 7^1\) (middle terms cancel)
- Factor: \(T(7-1) = 7^{n+1} - 7^1\)
- Divide: \(T = \frac{7^{n+1} - 7^1}{6}\)
This trick works even if every term has a constant coefficient.
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Back | Use the multiply-subract trick.<br><ol><li>Mutliply the series by its base: \(7T\)</li><li>Subtract: \(7T - T = 7^{n+1} - 7^1\) (middle terms cancel)</li><li>Factor: \(T(7-1) = 7^{n+1} - 7^1\)</li><li>Divide: \(T = \frac{7^{n+1} - 7^1}{6}\)</li></ol> | Use the multiply-subract trick.<br><ol><li>Mutliply the series by its base: \(7T\)</li><li>Subtract: \(7T - T = 7^{n+1} - 7^1\) (middle terms cancel)</li><li>Factor: \(T(7-1) = 7^{n+1} - 7^1\)</li><li>Divide: \(T = \frac{7^{n+1} - 7^1}{6}\)</li></ol><div>This trick works even if every term has a constant coefficient.</div> |