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# Help with proof by induction needed watch

1. I really get stuck with divisibility questions like this:

Prove by induction that for N for all positive integers,

f(n) = 2^(2n-1) + 3^(2n-1) is divisible by 5

I'm stuck on the induction step so

f(k+1) - f(k) = 2^(2k+1) +3^(2k+1) - 2^(2k-1) - 3^(2k-1)

Thanks for the help
2. 2^(n+2)=2^2 * 2^n=4(2^n).

Can u use this information to rewrite some terms of your final line so u can simplify?
I really get stuck with divisibility questions like this:

Prove by induction that for N for all positive integers,

f(n) = 2^(2n-1) + 3^(2n-1) is divisible by 5

I'm stuck on the induction step so

f(k+1) - f(k) = 2^(2k+1) +3^(2k+1) - 2^(2k-1) - 3^(2k-1)

Thanks for the help
^^^^^^.

Sorry forgot to quote u😂
4. (Original post by Shaanv)
^^^^^^.

Sorry forgot to quote u😂
alright. so I've got this now:

2(2^2k) + 3(3^2k) - 2^(k-1) - 3^(2k-1)

2(2^2k) + 3(3^2k) - 2^2k(1/2) - 3^2k(1/3)

and I'm lost again >_<
alright. so I've got this now:

2(2^2k) + 3(3^2k) - 2^(k-1) - 3^(2k-1)

2(2^2k) + 3(3^2k) - 2^2k(1/2) - 3^2k(1/3)

and I'm lost again >_<
Try getting everything as either 2^(2k-1) and 3^(2k-1).

Also i think i would try f(k)+f(k+1) for this proof by induction
I really get stuck with divisibility questions like this:

Prove by induction that for N for all positive integers,

f(n) = 2^(2n-1) + 3^(2n-1) is divisible by 5

I'm stuck on the induction step so

f(k+1) - f(k) = 2^(2k+1) +3^(2k+1) - 2^(2k-1) - 3^(2k-1)

Thanks for the help
Might be clearer if you just say

Then just slide the other terms into each brackets and subtract them to get:

And finish it off from there using the assumption that is div. by 5
7. (Original post by RDKGames)
Might be clearer if you just say
From here I think it's simpler to calculate f(k+1) - f(k); you get

and it's easy to split this into a mutiiple of f(k) (therefore divisible by 5) and an explicit multiple of 5.

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