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With a given example

$$ a_{n-1} = ca_{n-2} $$

general solution:

$$ a_{n} = c . c . a_{n-2} $$
$$ = c . c . a_{n-3} $$
$$ = c^n a_0 $$

Question: Find the general solution for the homogeneous equation $$ a_{n} = 5a_{n-1} $$

General solution:

$$ a_{n} = 5^n a_{0} $$

Is my general solution correct based on the given example?

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I believe you have a typo in the first equation, should it be $a_n = c \cdot a_{n-1}$? – Nicolas Villanueva May 5 '11 at 7:59
Yes. There is a typo in the first displayed formula, you meant $a_n=c\cdot a_{n-1}$, and in the third displayed line, you meant $a_n=c\cdot c\cdot c\cdot a_{n-3}$. And I would prefer to go forwards than backwards, $a_1=5a_0$, $a_2=5a_1=5^2a_0$, and so on. But backwards is fine. – André Nicolas May 5 '11 at 8:00
@Nicolas Villanueva: I have corrected the typo. @yunone: I am new to mathse. Thanks for the info. I will accept the answers – youcanlearnanything May 5 '11 at 8:10
@liangneh: The correction of the first typo should have been to $a_n=c\cdot a_{n-1}$. The correction you made is correct, but not as helpful. The next typo, on the third displayed line, remains uncorrected as of now. – André Nicolas May 5 '11 at 8:37
up vote 1 down vote accepted

Even if that first equation is a typo, that is the correct general solution. If a proof is necessary, I suggest using Induction on $n$.

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That was a given example for me to answer the question below. – youcanlearnanything May 5 '11 at 8:17
this is not a proving question – youcanlearnanything May 5 '11 at 9:59

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