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November 27, 2014

November 27, 2014

Posted by **GEMMA** on Wednesday, October 31, 2012 at 7:53am.

- maths - error -
**Steve**, Wednesday, October 31, 2012 at 11:51amAs written, it's obviously false. Try n=2

Did you mean 7n^2+4n+1? Nope; false for n=2

Got some error here

- maths -
**GEMMA**, Wednesday, October 31, 2012 at 12:57pmits 7^n+4^n+1

- maths -
**GEMMA**, Wednesday, October 31, 2012 at 1:11pmSteve

- maths -
**Steve**, Wednesday, October 31, 2012 at 4:37pmAh; that's a lot nicer.

it's true for n=1.

So, assume it's true for n=k.

7^(k+1) + 4^(k+1) + 1

= 7*7^k + 4*4^k + 1

= (1+6)*7^k + (1+3)*4^k + 1

= (7^k+4^k+1) + 6*7^k + 3*4^k

obviously, 6*7^k is divisible by 6

4^k is even, so 3*4^k is divisible by 6

So, since we're adding three items which are all multiples of 6, the whole is a multiple of 6.

Thus, the induction step holds.

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