Hello!
I am really not sure where to post this problem, so here we go.
Prove by mathematical induction:
\(\displaystyle P(n)\, \equiv \, \left(3 \cdot 5^{2n+1}\right)\, +\, 2^{3n+1}\, \Rightarrow \, P(n) \vdots 17\)
I have been trying to solve this problem and, at this moment, I can say I have no idea what to do in order to prove above equation is always divisible by 17. I tried, I really did, but I am simply stuck. If someone could help me out, at least a little, I would be much appreciated!
Edit:
This is my attempt:
P(1) is correct.
P(k) assume correct.
Prove for P(k+1) =>
3 * 52(k+1)+1 + 23(k+1)+1 =
=3 * 52 * 52k+1 + 23k+1 * 23 = 75 * 52k+1 + 23k+1 * 8 =
=(17*4)52k+1 + 7*52k+1 + (7+1)23k+1 =
=(17*4)52k+1 + 7(52k+1 + 23k+1) + 23k+1
That would be the closest. Leaves me with extra 23k+1. What am I missing?
I am really not sure where to post this problem, so here we go.
Prove by mathematical induction:
\(\displaystyle P(n)\, \equiv \, \left(3 \cdot 5^{2n+1}\right)\, +\, 2^{3n+1}\, \Rightarrow \, P(n) \vdots 17\)
I have been trying to solve this problem and, at this moment, I can say I have no idea what to do in order to prove above equation is always divisible by 17. I tried, I really did, but I am simply stuck. If someone could help me out, at least a little, I would be much appreciated!
Edit:
This is my attempt:
P(1) is correct.
P(k) assume correct.
Prove for P(k+1) =>
3 * 52(k+1)+1 + 23(k+1)+1 =
=3 * 52 * 52k+1 + 23k+1 * 23 = 75 * 52k+1 + 23k+1 * 8 =
=(17*4)52k+1 + 7*52k+1 + (7+1)23k+1 =
=(17*4)52k+1 + 7(52k+1 + 23k+1) + 23k+1
That would be the closest. Leaves me with extra 23k+1. What am I missing?
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