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sHOW THAT 5^2n -2^3n is divisible by 17

 Sep 29, 2018

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 #1
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\(\text{let }f(n) = 5^{2n} - 2^{3n} =25^n - 8^n\\ f(1) = 25-8 = 17 \text{ which is clearly divisible by 17}\)

 

\(\text{Now assume that }f(n) \text{ is divisible by 17. We need to show that }f(n+1) \text{ is as well}\\ f(n+1) = 25^{n+1} - 8^{n+1} = 25\cdot 25^n - 8\cdot 8^n = \\ 25\cdot 25^n-25\cdot 8^n + 17\cdot 8^n =\\ 25(25^n - 8^n) + 17\cdot 8^n =\\ 25 \cdot 17m + 17\cdot 8^n, (\text{ f(n) =17m by assumption} ) \\ 17(25m + 8^n) \\ \text{and this is clearly divisible by 17}\)

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 Sep 29, 2018
 #1
avatar+6244 
+3
Best Answer

\(\text{let }f(n) = 5^{2n} - 2^{3n} =25^n - 8^n\\ f(1) = 25-8 = 17 \text{ which is clearly divisible by 17}\)

 

\(\text{Now assume that }f(n) \text{ is divisible by 17. We need to show that }f(n+1) \text{ is as well}\\ f(n+1) = 25^{n+1} - 8^{n+1} = 25\cdot 25^n - 8\cdot 8^n = \\ 25\cdot 25^n-25\cdot 8^n + 17\cdot 8^n =\\ 25(25^n - 8^n) + 17\cdot 8^n =\\ 25 \cdot 17m + 17\cdot 8^n, (\text{ f(n) =17m by assumption} ) \\ 17(25m + 8^n) \\ \text{and this is clearly divisible by 17}\)

Rom Sep 29, 2018

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