Loser66
  • Loser66
Given \(a_n = 2^n + 5*3^n\) for n =0,1,2,.... Prove that \(a_n = 5a_{n-1} -6a_{n-2} \) for \(n\geq 2\) Please, help
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
i have no idea what happens if you try it directly?
Loser66
  • Loser66
What do you mean? I try induction, but I don't get the correct solution
anonymous
  • anonymous
i mean just write out what \[5a_{n-1} -6a_{n-2}\] is

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anonymous
  • anonymous
it is not an induction proof i don't think
anonymous
  • anonymous
in other words, write out \[5a_{n-1} -6a_{n-2}=5\left(2^{n-1}+5\times 3^{n-1}\right)-6\left(2^{n-2}+5\times 3^{n-2}\right)\]
anonymous
  • anonymous
this is my best guess then muck around with the exponents
Loser66
  • Loser66
Ok, let me try .
anonymous
  • anonymous
or as we say in elementary algebra, multiply out using the distributive property, combine like terms i think it should work
ganeshie8
  • ganeshie8
To simplify the algebra, you may use this trick : The given recurrence relation is linear. So, if \({r_1}^n\) and \({r_2}^n\) are two solutions, then by linearity it follows that all the linear combinations of those solutions are also solutions.
Loser66
  • Loser66
Can I use induction? But it doesn't work though. :(
Empty
  • Empty
\[a_n = 5a_{n-1}-6a_{n-2}\] Plug 'em in: \[2^n+5*3^n = 5(2^{n-1}+5*3^{n-1}) - 6(2^{n-1}+5*3^{n-1})\]Rearrange the right hand side: \[2^n+5*3^n = \left( \frac{5}{2} - \frac{6}{4} \right) 2^n + \left( \frac{25}{3} - \frac{30}{9} \right) 3^n\]
ganeshie8
  • ganeshie8
Induction works nicely tooo
Loser66
  • Loser66
I got it. Thank you so much.
Loser66
  • Loser66
@Empty
Loser66
  • Loser66
@ganeshie8 Can you show me induction? Please
Empty
  • Empty
Whoops, when I copied and pasted I just realized I didn't put in the -2 in the exponent I left it as -1 oh well cool glad it helped
ganeshie8
  • ganeshie8
By pluggin in, or by induction, you're just doing more of old highschool algebra.... you're not learning anything new...
Loser66
  • Loser66
I know and this is not my problem. I got mad when I can't find out the mistake I have on solving it for...... a high school student. But now I got what I missed. :)
ganeshie8
  • ganeshie8
Base case : \(a_0=6,a_1=17,a_2=49\) and \(49 = 5*17-6*6\) \(\checkmark\) Induction hypothesis : Assume \(a_n = 5a_{n-1}-6a_{n-2}\) is satisfied for the terms \(n=k,k-1\). Induction step : \(a_{k+1} = 2^{k+1}+5*3^{k+1}\\~\\ =2*2^k + 15*3^k\\~\\ =(5-3)*2^k+(25-10)*3^k\\~\\ =5*2^k+25*3^k- (3*2^k+10*3^k)\\~\\ =5(2^k+5*3^k) - 6(2^{k-1}+5*3^{k-1})\\~\\ =5a_k - 6a_{k-1}\\~\\ \blacksquare \)
Loser66
  • Loser66
Thank you very much.

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