MHB How Do You Prove a Recursive Sequence Using Strong Induction?

AI Thread Summary
The discussion focuses on proving a recursive sequence defined by f(0) = 0, f(1) = 1, and f(n+1) = 3f(n) + 10f(n-1) using strong induction to show that f(n) < 5^n for all integers n ≥ 0. The base case is verified for n = 0 and n = 1, establishing that f(0) < 5^0 and f(1) < 5^1. The induction step involves demonstrating that if f(n) < 5^n and f(n-1) < 5^(n-1), then f(n+1) can be shown to be less than 5^(n+1) through algebraic manipulation. Clarifications were requested regarding the arithmetic steps taken to reach the conclusion, which were subsequently explained, making the process clearer for participants. The conversation highlights common challenges with strong induction and the importance of clear arithmetic in proofs.
andrew1
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Hi,

I'm currently having trouble with using strong induction to prove a recursive sequence and don't know where to begin, any help would be great thanks.

Define a recursive sequence f(0), f(1), f(2),... by
f(0) = 0
f(1) = 1
f(n+1) = 3f(n) + 10f(n-1), for all integers n>=1

Prove by strong induction that f(n) < 5^n for all integers n>=0
 
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andrew said:
Hi,

I'm currently having trouble with using strong induction to prove a recursive sequence and don't know where to begin, any help would be great thanks.

Define a recursive sequence f(0), f(1), f(2),... by
f(0) = 0
f(1) = 1
f(n+1) = 3f(n) + 10f(n-1), for all integers n>=1

Prove by strong induction that f(n) < 5^n for all integers n>=0

because f(1) < $5^1$ and f(0) = 0 <5 $^0$ base step is true

now we need to prove the induction step

$f(n ) < 5^n$
$f(n-1 ) < 5^{n-1}$
f(n+2) = 3 f(n) + 10 f(n-1)

< $3 * 5^n + 10 * 5^{n-1}$
< $3 * 5^n + 2 * 5^n$
< $5 * 5^n $
< $5^{n+1}$

so if it true for n and n-1 it is true for n+ 1
so induction step is proved
hence proved
 
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Thank you, but I'm confused as to how you arrived at that conclusion, could you provide a description of the process that you went through for each of the steps please.
 
andrew said:
Thank you, but I'm confused as to how you arrived at that conclusion, could you provide a description of the process that you went through for each of the steps please.
can you explain in which step there was confusion.
 
kaliprasad said:
can you explain in which step there was confusion.

This part doesn't make much sense to me

< $3 * 5^n + 10 * 5^{n-1}$
< $3 * 5^n + 2 * 5^n$
< $5 * 5^n $
< $5^{n+1}$
 
andrew said:
This part doesn't make much sense to me

< $3 * 5^n + 10 * 5^{n-1}$
< $3 * 5^n + 2 * 5^n$
< $5 * 5^n $
< $5^{n+1}$

OK \:
we have f(n) < $5^n$so 3 f( n) < 3 * 5^n

f(n-1) < $5^{n-1}$ it is true upto n so for all lower values

so f(n+1) = 3f(n) + 10 f(n-1)
< $3 * 5^n + 10 * 5^{n-1}$ (adding last 2)
< $3 * 5^n + 2 * 5 * 5^{n-1}$ (as 10 -= 2 * 5)
< $3 * 5^n + 2 * 5^n$
< $(3 + 2) * 5^n$
< $5 * 5^n$
< $5^{n+1}$

I hope now all steps are clear. I have just put the values and did arithmetic
 
kaliprasad said:
OK \:
we have f(n) < $5^n$so 3 f( n) < 3 * 5^n

f(n-1) < $5^{n-1}$ it is true upto n so for all lower values

so f(n+1) = 3f(n) + 10 f(n-1)
< $3 * 5^n + 10 * 5^{n-1}$ (adding last 2)
< $3 * 5^n + 2 * 5 * 5^{n-1}$ (as 10 -= 2 * 5)
< $3 * 5^n + 2 * 5^n$
< $(3 + 2) * 5^n$
< $5 * 5^n$
< $5^{n+1}$

I hope now all steps are clear. I have just put the values and did arithmetic

Hmmk, makes more sense now, strong induction always seems to confuse me :(
 
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