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I'm not able to find particular solution of
$a_n-2a_{n-1}$=$3*2^n$
What i've tried

  1. Given RR is $a_n-2a_{n-1}$=$3*2^n$
  2. For the particular solution observe the r.h.s of the equation(1)
  3. It is $3*2^n$=(a constant)*$2^n$
  4. Consider the P.S =(a constant)*$2^n$
  5. $a_n^{(p)}$=A*$2^n$
  6. $a_{n-1}$=A*$2^{n-1}$
  7. Substituting this value in Eq..(1)
  8. A*$2^n-2$A*$2^{n-1}=$$3*2^n$
  9. $A*2^n(1-1)=3*2^n$
  10. $A*2^n0=3*2^n$
  11. This were i'm stuck i'm not getting $A$ value Please help..:(
CY5
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    The problem come from the fact that $2$ is already the solution of the characteristic equation. Then the $\cdots n \times 2^n$ given in the answers. – Claude Leibovici Nov 01 '15 at 12:41
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    Hint: $$\frac{a_n}{2^n} = 3 + \frac{a_{n-1}}{2^{n-1}} = 6 + \frac{a_{n-2}}{2^{n-2}} = 9 + \frac{a_{n-3}}{2^{n-3}} = 12 + \frac{a_{n-4}}{2^{n-4}} = \ldots \ldots +\frac{a_{0}}{2^{0}}$$ – Did Nov 01 '15 at 12:55
  • Hitting the 80 character limit in the middle of a fraction made @Did's hint harder to parse than it should be. I edited it by adding a little bit of space. Consequently the 80 character limit was reached outside fractions, and allows the formula to render more or less correctly. Unfortunately that extra space now shows in the $n-3$ -term. Sorry, I can't do better. – Jyrki Lahtonen Nov 03 '15 at 18:01
  • @JyrkiLahtonen Well done, I am in awe of your teXnical ability... or is it some MathJax thing? Anyway, thanks. – Did Nov 03 '15 at 18:04
  • YW, @Did. Here is a meta thread with more information. The way the site software handles the comments means that every 80 characters some non-TeX space characters are inserted, and that can understandably confuse MathJaX, if those extra characters come at a bad time. – Jyrki Lahtonen Nov 03 '15 at 18:11

3 Answers3

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If you are looking for a particular solution, you could try $a_0=0$ which would give $$a_1=2\times 0 + 3\times 2^1 = 6$$ $$a_2=2\times 6 + 3\times 2^2 = 24$$ $$a_3=2\times 24 + 3\times 2^2 = 72$$ $$\cdots$$ and see by inspection and prove by induction that $a_n=3n \, 2^n$.

A similar process of inspection and induction would give the general solution $a_n=(3n+a_0)2^n$.

Henry
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In these "resonant" type cases, try solutions of the form $$a_n \propto n\lambda^n$$

Note that solutions to $a_n - 2a_{n-1} = 0$ can be added to the solution to give further solutions.

Chris Kerridge
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Let $a_m=b_m+(cm+d)2^m$

$$3\cdot2^n=b_n+(cn+d)2^n-2\{b_{n-1}+(cn-c+d)2^{n-1}\}=b_n-2b_{n-1}+c2^n$$

Choose $c=3$

lab bhattacharjee
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