A sequence Ln is such that L1 = 0.2, L2 = 0.1 and

I am not sure how to use mathcad for this, so far i did this in mathcad

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## closed as off topic by Austin Mohr, Thomas, Chris Eagle, Norbert, tomaszOct 6 '12 at 11:49

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Let $$A=\begin{pmatrix} 2&-1.4\\1&0 \end{pmatrix}$$ and let
$$v_0=\begin{pmatrix} 0.1\\0.2 \end{pmatrix}$$ You will get the vector $$\begin{pmatrix} L_n\\L_{n-1} \end{pmatrix}$$ by multiplying $$\begin{pmatrix} L_{n-1}\\L_{n-2} \end{pmatrix}$$ by $A$. So you will want to multiply $v_0$ by a high power of $A$. The program is good at matrix manipulations.
 hey, can this be done in office excel as well? just a tot my friend put in to my brain – JackyBoi Oct 3 '12 at 6:01 Don't know. Only have used that for keeping track of grades. Are you sure about your initial conditions? One gets negative numbers. That's OK, but makes me wonder whether there may be a typo. – André Nicolas Oct 3 '12 at 6:04 At where specifcally? that is an image so perhaps it might look like that, and honestly this is a Fibonnaci sequence right? and I really not sure how to get the answer.. even after your explanation – JackyBoi Oct 3 '12 at 6:05 You can ask the excel part in stackoverflow.com – Pragabhava Oct 3 '12 at 6:06 Well, it is a linear recurrence with constant coefficients, but very different from the Fibonacci recurrence $F_n=F_{n-1}+F_{n-2}$. As to MathCad code, perhaps someone will supply it. There are also techniques from Linear Algebra for doing such computations fast, but that might take us too far afield. Certainly you could alternately just program the recurrence. – André Nicolas Oct 3 '12 at 6:09