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AravindG

  • 4 years ago

justify (5^k-5) always divisible by 4

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  1. OsmondF
    • 4 years ago
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    |dw:1332390749082:dw|

  2. Ishaan94
    • 4 years ago
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    Use Induction!

  3. NotTim
    • 4 years ago
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    Logs? YOu try logarthmic? WHY DID YOU ASK ME...

  4. AravindG
    • 4 years ago
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    how??

  5. Ishaan94
    • 4 years ago
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    Hmm \[\frac{(1+4)^k - 1 - 4}{4}\] Try to use the binomial expansion, the 1 would cancel, leaving only the factors of 4.

  6. AravindG
    • 4 years ago
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    srry i didnt get that

  7. AravindG
    • 4 years ago
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    what to do with (1+4)^k

  8. vishal_kothari
    • 4 years ago
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    |dw:1332391543618:dw||dw:1332391574845:dw||dw:1332391595432:dw||dw:1332391616609:dw||dw:1332391646332:dw||dw:1332391674792:dw|

  9. Ishaan94
    • 4 years ago
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    Use the binomial expansion \[(1+4)^k = \underbrace{\binom{k}{0}4^{0}1^{k}}_{\text{This is 1}} + \binom{k}{1}4^1 1^{k-1} + \ldots + \binom{k}{k}4^k \] \[\frac{1 + \binom{k}{1}4^1 1^{k-1} + \ldots + \binom{k}{k}4^k -1 -4}{4}\]

  10. AravindG
    • 4 years ago
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    oh k thx evryone

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