AravindG justify (5^k-5) always divisible by 4 one year ago one year ago

1. OsmondF

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2. Ishaan94

Use Induction!

3. NotTim

Logs? YOu try logarthmic? WHY DID YOU ASK ME...

4. AravindG

how??

5. Ishaan94

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

srry i didnt get that

7. AravindG

what to do with (1+4)^k

8. vishal_kothari

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9. Ishaan94

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

oh k thx evryone