+26393 f(1)=20 , f(n)=3*f(n-1)-60 find the 10th term
\(\small{ \begin{array}{lrcl} & f_n &=& 3\cdot f_{n-1} - 60 \qquad a=3 \qquad c=-60\\ \text{or general}: & f_n &=& a\cdot f_{n-1} + c \\ \hline \\ & f_1 &=& a\cdot f_0 + c \\\\ & f_2 &=& a\cdot f_1 + c \\ & f_2 &=& a\cdot [a\cdot f_0 + c] + c \\ & f_2 &=& a^2\cdot f_0 + a\cdot c + c \\ & f_2 &=& a^2\cdot f_0 + c\cdot(1+a) \\\\ & f_3 &=& a\cdot f_2 + c \\ & f_3 &=& a\cdot [a^2\cdot f_0 + c\cdot(1+a)] + c \\ & f_3 &=& a^3\cdot f_0 + a\cdot c\cdot(1+a) + c \\ & f_3 &=& a^3\cdot f_0 + c\cdot(a+a^2) + c \\ & f_3 &=& a^3\cdot f_0 + c\cdot(1+a+a^2) \\\\ & f_4 &=& a\cdot f_3 + c \\ & f_4 &=& a\cdot [a^3\cdot f_0 + c\cdot(1+a+a^2)] + c \\ & f_4 &=& a^4\cdot f_0 + a\cdot c\cdot(1+a+a^2) + c \\ & f_4 &=& a^4\cdot f_0 + c\cdot(a+a^2+a^3) + c \\ & f_4 &=& a^4\cdot f_0 + c\cdot(1+a+a^2+a^3) \\\\ \hline \text{general solution}: & f_n &=& a^n\cdot f_0 + c\cdot (1+a+a^2+a^3+...+ a^{n-1}) \\ & && 1+a+a^2+a^3+...+ a^{n-1} = \frac{1-a^n}{1-a}\\ & f_n &=& a^n\cdot f_0 + c\cdot \frac{1-a^n}{1-a} \\ & f_n &=& a^n\cdot f_0 + c\cdot \left( \frac{1}{1-a} - \frac{a^n}{1-a} \right) \\ & f_n &=& a^n\cdot f_0 + c\cdot\frac{1}{1-a} - c\cdot\frac{a^n}{1-a} \\ & f_n &=& a^n\cdot f_0 - c\cdot\frac{a^n}{1-a} + c\cdot\frac{1}{1-a} \\ & && \boxed{~ \begin{array}{lrcl} & f_n &=& a^n\cdot \left( f_0 - c\cdot\frac{1}{1-a} \right) + c\cdot\frac{1}{1-a} \\ & f_n &=& x + \beta \cdot a^n \\ &&& x = c\cdot\frac{1}{1-a} \\ &&& \beta = f_0 - c\cdot\frac{1}{1-a} \\ &&& \beta = f_0 - x \end{array} ~}\\ \hline \\ \text{general}: & f_n &=& a\cdot f_{n-1} + c \\ \text{general solution}: & f_n &=& x + \beta \cdot a^n \qquad x = \frac{c}{1-a} \qquad \beta = f_0 - x \\\\ & f_n &=& 3\cdot f_{n-1} -60 \qquad a=3 \qquad c=-60\\\\ & f_1 &=& 3 \cdot f_0 - 60 \\ & 3 \cdot f_0 &=& f_1 + 60 \\ & f_0 &=& \frac{ f_1 + 60 } {3} \qquad f_1 = 20\\ & f_0 &=& \frac{ 20 + 60 } {3} \\ & f_0 &=& \frac{ 80 } {3} \\\\ & x &=& \frac{c}{1-a} \\ & x &=& \frac{-60}{1-3} \\ & x &=& \frac{-60}{-2} \\ & x &=& 30 \\ \\ & \beta &=& f_0 - x \\ & \beta &=& \frac{ 80 } {3} - 30 \\ & \beta &=& -\frac{ 10 } {3} \\\\ & f_n &=& 30 -\frac{ 10 } {3} \cdot a^n \\ & f_{10} &=& 30 -\frac{ 10 } {3} \cdot 3^{10} \\ & f_{10} &=& 30 -\frac{ 10 } {3} \cdot 59049 \\ & f_{10} &=& 30 -196830\\ & \mathbf{f_{10} }& \mathbf{=} &\mathbf { -196800 } \end{array} }\)
+129852 f(1) = 20
f(n)=3*f(n-1)-60
Using WolframAlpha, the generalized "formula" for the "nth" term is :
f(n) = (-10/3) (3n - 9)
So
f(10) = (-10/3) (310 - 9) = - 196,800
+250 We have \(f_n = 3f_{n-1}-60\) and \(f(1)=f_1=20\)
For a difference equation in the form \(f_n = af_{n-1}+c\) , the general solution takes form of: \(f_n=x+\beta a^n\) where x is the fixed point of the difference equation:
1) \(f(0)=f_0 = \frac{f_1+60}{3}=\frac{80}{3}\)
2) Finding the fixed point \(x\) -> Let \(f_n=f_{n-1}=f\), then:
\(f=3f+60 \quad=>\quad f=30\) so the fixed point \(x\) is 30.
3) To find \(\beta\) in the general solution, work out the case n=0:
\(f_0=\frac{80}{3}=30 +\beta 3^0 = 30+\beta*1 \quad=>\quad \beta=\frac{80}{3}-30=\frac{-10}{3}\)
4) Now we have the general solution just plug in whatever term you want to find, in this case the 10th term:
\(f_{n}=30-\frac{10}{3}*3^{n}\\f_{10}=30-\frac{10}{3}*3^{10} = -196800\)
+26393 f(1)=20 , f(n)=3*f(n-1)-60 find the 10th term
\(\small{ \begin{array}{lrcl} & f_n &=& 3\cdot f_{n-1} - 60 \qquad a=3 \qquad c=-60\\ \text{or general}: & f_n &=& a\cdot f_{n-1} + c \\ \hline \\ & f_1 &=& a\cdot f_0 + c \\\\ & f_2 &=& a\cdot f_1 + c \\ & f_2 &=& a\cdot [a\cdot f_0 + c] + c \\ & f_2 &=& a^2\cdot f_0 + a\cdot c + c \\ & f_2 &=& a^2\cdot f_0 + c\cdot(1+a) \\\\ & f_3 &=& a\cdot f_2 + c \\ & f_3 &=& a\cdot [a^2\cdot f_0 + c\cdot(1+a)] + c \\ & f_3 &=& a^3\cdot f_0 + a\cdot c\cdot(1+a) + c \\ & f_3 &=& a^3\cdot f_0 + c\cdot(a+a^2) + c \\ & f_3 &=& a^3\cdot f_0 + c\cdot(1+a+a^2) \\\\ & f_4 &=& a\cdot f_3 + c \\ & f_4 &=& a\cdot [a^3\cdot f_0 + c\cdot(1+a+a^2)] + c \\ & f_4 &=& a^4\cdot f_0 + a\cdot c\cdot(1+a+a^2) + c \\ & f_4 &=& a^4\cdot f_0 + c\cdot(a+a^2+a^3) + c \\ & f_4 &=& a^4\cdot f_0 + c\cdot(1+a+a^2+a^3) \\\\ \hline \text{general solution}: & f_n &=& a^n\cdot f_0 + c\cdot (1+a+a^2+a^3+...+ a^{n-1}) \\ & && 1+a+a^2+a^3+...+ a^{n-1} = \frac{1-a^n}{1-a}\\ & f_n &=& a^n\cdot f_0 + c\cdot \frac{1-a^n}{1-a} \\ & f_n &=& a^n\cdot f_0 + c\cdot \left( \frac{1}{1-a} - \frac{a^n}{1-a} \right) \\ & f_n &=& a^n\cdot f_0 + c\cdot\frac{1}{1-a} - c\cdot\frac{a^n}{1-a} \\ & f_n &=& a^n\cdot f_0 - c\cdot\frac{a^n}{1-a} + c\cdot\frac{1}{1-a} \\ & && \boxed{~ \begin{array}{lrcl} & f_n &=& a^n\cdot \left( f_0 - c\cdot\frac{1}{1-a} \right) + c\cdot\frac{1}{1-a} \\ & f_n &=& x + \beta \cdot a^n \\ &&& x = c\cdot\frac{1}{1-a} \\ &&& \beta = f_0 - c\cdot\frac{1}{1-a} \\ &&& \beta = f_0 - x \end{array} ~}\\ \hline \\ \text{general}: & f_n &=& a\cdot f_{n-1} + c \\ \text{general solution}: & f_n &=& x + \beta \cdot a^n \qquad x = \frac{c}{1-a} \qquad \beta = f_0 - x \\\\ & f_n &=& 3\cdot f_{n-1} -60 \qquad a=3 \qquad c=-60\\\\ & f_1 &=& 3 \cdot f_0 - 60 \\ & 3 \cdot f_0 &=& f_1 + 60 \\ & f_0 &=& \frac{ f_1 + 60 } {3} \qquad f_1 = 20\\ & f_0 &=& \frac{ 20 + 60 } {3} \\ & f_0 &=& \frac{ 80 } {3} \\\\ & x &=& \frac{c}{1-a} \\ & x &=& \frac{-60}{1-3} \\ & x &=& \frac{-60}{-2} \\ & x &=& 30 \\ \\ & \beta &=& f_0 - x \\ & \beta &=& \frac{ 80 } {3} - 30 \\ & \beta &=& -\frac{ 10 } {3} \\\\ & f_n &=& 30 -\frac{ 10 } {3} \cdot a^n \\ & f_{10} &=& 30 -\frac{ 10 } {3} \cdot 3^{10} \\ & f_{10} &=& 30 -\frac{ 10 } {3} \cdot 59049 \\ & f_{10} &=& 30 -196830\\ & \mathbf{f_{10} }& \mathbf{=} &\mathbf { -196800 } \end{array} }\)
+250 Not a problem! The "fixed point" refers to the value(s) of \(f_{n-1}\) that when you work out \(f_n\), you get the same value that \(f_{n-1}\) was, this putting you in a never ending loop and hence is a fixed point (or equilibrium).
Here's the method for finding the fixed point (explained pretty loosely):
-> By definition the fixed point means that the next number should be the same as the current one which is equivalent to saying \(...=f_{n-2}=f_{n-1}=f_n=f_{n+1}=...\) This means we can forget about the subscripts and treat everything as the same unknown \(f\).
-> This means at the fixed point the equation \(f_n=3f_{n-1}-60\) transforms into \(f=3f-60\).
-> Now you just solve for \(f\) and that will be your fixed point
In this example, \(f = 30\) and substituting it into the original difference equation we get:
\(f_{n-1}=30\\f_n=3f_{n-1}-60\quad=>\quad f_n=3(30)-60=30\quad=>\quad f_n=f_{n-1}\)
so this is a fixed point.
+129852 OK, Brodudedoodebrodude............got it!!!!........Thanks!!!!
[Shouldn't the " + " sign in Step 2 of your original answer be a " -" instead??? ]
+250 Totally should be a minus, my bad. Glad I could help =)
Edit: Seems I cant edit my original post =/
+129852 Thanks , again.....I haven't seen this before and the "fixed point" concept is a new one on me......your explanation was excellent....!!!!!
[BTW......I think Melody mentioned to me that the "new and improved" forum will not let you edit LaTex......!!!!........bummer....!!!! ]
+129852 Thanks to you too, heureka........that's two methods that I have to learn from......!!!!!
+118673 Hi Brodude,
Welcome back to the forum, it is great to see you again
I have had a lot of problems with editing LaTex on the new forum but I just went back into an old post and edited the LaTex.
So you can edit it some of the time at least.
It was very unresponsive. I had to click it many times to get the yellow box to turn blue. It has to go blue before you have any chance of editing.
Anyway, if you are persistent enough maybe you will be able to edit your LaTex too.
GOOD LUCK !!
+118673 f(1)=20 , f(n)=3*f(n-1)-60 find the 10th term
I have been thinking about this question ://
I would do it the same as Heureka but I was just thinking about why this sequence is so well known.
It is important because it is the same logic that would be used for future value of an ordinary annutiity with an original deposit.
Say you start a bank account off with P dollars. Then you add M dollars at the end of every month
The interest rate is 100r% per month that is r as a decimal. Each month the amount will grow by a factor of 1+r which I will call R R=1+r
Beginning of month 1 F0 = P
End of month 1 F1= PR+M
End of month 2 F2 = R*F1 +M
F2 = R(PR+M)+M
F2 = PR^2+MR+M
End of month 3 F3 = F2 *R +M so F(n) = R*F(n-1)+M
F3 = (PR^2+MR+M) *R +M
F3 = PR^3 + MR^2 + MR +M
F3 = PR^3 + M(R^2 + R +1)
--
End of nth month Fn = PR^n + M(R^(n-1) + R^(n-2)...............+R +1)
This is a GP a=1, r=R, n=n
(R^(n-1) + R^(n-2)...............+R +1)
\(s_n=\frac{1(R^n-1)}{R-1}\\ s_n=\frac{R^n-1}{R-1}\\\)
\(\mbox{End of the nth month= }PR^n+\frac{M(R^n-1)}{R-1}\)
this is what your money will have accumulated to:
--------------------------
Relating this to the given problem - which is not a money problem but it is of the same form as an annuity.
f(1)=20 , f(n)=3*f(n-1)-60 find the 9th term (It is the 9th because it starts at 1 instead of at 0)
P=20 R=3 M=-60 n=9
\(\mbox{End of the nth period= }T_n= PR^n+\frac{M(R^n-1)}{R-1}\\ \mbox{End of the 9th period= }T_{9} =20*3^{9}+\frac{-60(3^9-1)}{3-1}\\ \mbox{End of the 9th period= }T_{9} =20*3^{9}+\frac{-60(3^9-1)}{3-1}\\ \mbox{End of the 9th period= }T_{9} =20*3^{9}-30(3^9-1)\\ \mbox{End of the 9th period= }T_{9} = -196,800\\\)
