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Can any one help me to solve this $$ U_{n+1}=5U_n-6U_{n-1} $$
$$ U_0=1,U_1=4 $$ $$ V_n=U_{n+1}-2U_n $$ I've proved that $V_n$ is a geometric sequence and I managed to write $V_n$ by $n$ $$ V_n=2.3^n $$ How can I write $U_n$ formula by $n$?
If any one can help me I'll be thankful.

1 Answers 1

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Hint

Call $$U_{n+1}=\lambda^n$$

and get

$$\lambda^2-5\lambda+6=0$$

conclude that

$$U_n=a\cdot2^n+b\cdot3^n$$

and find $a$ and $b$.

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    Thanks, but how did u get that $U_n=a.2ⁿ+b.3ⁿ$ – 2017-01-23
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    Intact I just couldn't figure out what It done , Thanks anyway :) – 2017-01-23
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    See that $\lambda=2$ or $\lambda=3$ and its linear combination also fit the recurrence. – 2017-01-23