Induction help!!!! (1 Viewer)

[teanido]

Show 3^a + 7^a is divisible by 10 for all odd integers of a...

 

[pyrodude1031]

i will do my best to type in the powers ... they will be indicated using the ^ button followed by the round brakets() where necessary.
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Step One : If n=1 3^1+7^1 = 10 which is dividisble by 1. Hence statement is true for n=1.

Step two: Assume statement is true for n=k. i.e. 3^k + 7^k = 10M where M is any positive integer.
RTP statement is true for all postive odd integers i.e. statement is true for n=k+2.
i.e. RTP 3^(k+2) + 7^(k+2) = 10N where N is another positive integer.
LHS
=3^(k+2) + 7^(k+2)
=[3^2][3^k + 7^k] - [3^2][7^k] + 7^(k+2)
=10M[3^2] + [7^k][-9 + 49]
=10{[3^2]M + 4[7^k]}
=RHS

Step Three: the Bullshit that follows.. CBB type it up ...

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hope its clear enuff .. may be this can be done by taking out the 7^2 instead of the 3^2 . Same thing follows ...

Pyrodude1031 ...

 

[Affinity]

since a is odd, one has factorization:

3^a + 7^a = (3 + 7)(3^(a-1) - 3^(a-2)*7 + ... + 7^(a-1))
obviously divisible by 10