Queen_of_Random
New member
- Joined
- Oct 12, 2022
- Messages
- 3
We can't help you get unlost without some idea of where you are.Hi, I got quite lost with those ones:
View attachment 34313
Yes, it should beThe first one does not seem valid for a couple of reasons.
n=1, as a1 and a2 are given.What is the base case for the third problem?
I am addressing only your third problem, which is a bit tricky.
[math] \text {Given } a_1 = 5, \ a_2 = 13, \text { and } a_{n+2} = 5a_{n+1} - 6a_n \text { for any positive integer } n, \text {Prove } a_n = 2^n + 3^n. [/math]
What is tricky is figuring out how to start the induction. [imath]a_1 = 5 = 2^1 + 3^1[/imath] is necessary but trivial and does not help you with induction because [imath]a_{-1}[/imath] and [imath]a_0[/imath] are not defined.
Similarly, [imath]a_2 = 13 = 4 + 9 = 2^2 + 3^2[/imath] is necessary but trivial and does not help you with induction because [imath]a_0[/imath] is not defined.
So, we must start with [imath]a_3[/imath] as our base case for induction. It is not difficult if we see
[math]n = 3 \implies a_3 = a_{1 + 2} = 5a_{1+1} - 6a_{1} \implies \\ a_{1+2} = 5a_2 - 6a_1 = 65 - 30 = 35 = 8 + 27 = 2^3 + 3^3 \implies\\ a_{1+2} = 2^{(1+2)} + 3^{(1+2)}.[/math]
Now we can say
[math]a_1 = 2^1 + 3^1, \ a_2 = 2^2 + 3^2 =a_{1+1}, \text { and } a_3 = 2^3 + 4^3 = a_{1+2}.\\ \text {Let } k \text { be an ARBITRARY positive integer such that }\\ a_k = 2^k + 3^k, \ a_{k+1} = 2^{(k+1)} + 3^{(k+1)}, \text { and } a_{k+2} = 2^{(k+2)} + 3^{(k+2)}.[/math]
Now what?
Whether there are restrictions on n is not explicitly specified.Why can you start with [imath]a_3[/imath] but not with [imath]a_0[/imath]? After all, [imath]a_0 = \frac{5a_1-a_2}{6}[/imath].
I am a retired software engineer, so 0 seems like a natural first index to me And it's been a while since I've written anything in FortranI simply assumed that n was restricted to positive integers. Silly me.
I think JeffM is a retired banker → he does not like 0 as a starting number !!I am a retired software engineer, so 0 seems like a natural first index to me And it's been a while since I've written anything in Fortran
Jeff, I replied to your question earlier saying you start with n=1. Since the formula was a_(n+2)= 5a_(n+1) - 6a_n, you start with n=1, as that tell you what a_3 equals--which is the starting point.Whether there are restrictions on n is not explicitly specified.
Let‘s follow your hypothesis through and assume that n can be any integer.
[math] a_{0+2} = 5a_{0+1} - 6a_0 \implies a_2 = 5a_1 - 6a_0 \implies \\ a_0 = \dfrac{5a_1 - a_2}{6} = \dfrac{5 * 5 - 13}{6} = \dfrac{12}{6} = 2 = 1 + 1 = 2^0 + 3^0.\\ a_{-1+2} = 5a_{-1+1} - 6a_{-1} \implies a_1 = 5a_0 - 6a^{-1} \implies \\ a_{-1} = \dfrac{5a_0 - a_1}{6} = \dfrac{5 * 2 - 5}{6} = \dfrac{5}{6} = \dfrac{3}{6} + \dfrac{2}{6} = \dfrac{1}{2} + \dfrac{1}{3} = 2^{-1} + 3^{-1}. [/math]
I simply assumed that n was restricted to positive integers. Silly me.
No, I understood your point. I just was not sure the OP was with us.Jeff, I replied to your question earlier saying you start with n=1. Since the formula was a_(n+2)= 5a_(n+1) - 6a_n, you start with n=1, as that tell you what a_3 equals--which is the starting point.