Question

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

Answer 1

i have no idea
what happens if you try it directly?

Answer 2

What do you mean? I try induction, but I don't get the correct solution

Answer 3

i mean just write out what \[5a_{n-1} -6a_{n-2}\] is

Answer 4

it is not an induction proof i don't think

Answer 5

in other words, write out
\[5a_{n-1} -6a_{n-2}=5\left(2^{n-1}+5\times 3^{n-1}\right)-6\left(2^{n-2}+5\times 3^{n-2}\right)\]

Answer 6

this is my best guess
then muck around with the exponents

Answer 7

Ok, let me try .

Answer 8

or as we say in elementary algebra, multiply out using the distributive property, combine like terms
i think it should work

Answer 9

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.

Answer 10

Can I use induction? But it doesn't work though. :(

Answer 11

\[a_n = 5a_{n-1}-6a_{n-2}\] 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:
\[2^n+5*3^n = \left( \frac{5}{2} - \frac{6}{4} \right) 2^n + \left( \frac{25}{3} - \frac{30}{9} \right) 3^n\]

Answer 12

Induction works nicely tooo

Answer 13

I got it. Thank you so much.

Answer 14

@Empty

Answer 15

@ganeshie8 Can you show me induction? Please

Answer 16

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

Answer 17

By pluggin in, or by induction, you're just doing more of old highschool algebra.... you're not learning anything new...

Answer 18

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. :)

Answer 19

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 :
\(a_{k+1} = 2^{k+1}+5*3^{k+1}\\~\\
=2*2^k + 15*3^k\\~\\
=(5-3)*2^k+(25-10)*3^k\\~\\
=5*2^k+25*3^k- (3*2^k+10*3^k)\\~\\
=5(2^k+5*3^k) - 6(2^{k-1}+5*3^{k-1})\\~\\
=5a_k - 6a_{k-1}\\~\\
\blacksquare
\)

Answer 20

Thank you very much.