Solve an+2 = 4an+1 + 21an; n  0 and a0 = 3; a1 = 4

Given recurrence relation is-

$a_{n+2}=4a_{n+1}+21a_n$ Where, $a_o=3,a_1=4$

The characterstics equation of the above relation is given by-

$x^2=4x+21\\\Rightarrow x^2-4x-21=0\\ \Rightarrow (x-7)(x+3)=0$

Then the roots of equation are $7,-3$

So ,

$a_n=c_1(7)^n+c_2(-3)^n~~~~~-(1)$

At $n=0$

$a_o=c_1+c_1\Rightarrow c_1+c_2=3~~~~~-(2)$

At $n=1$

$a_1=7c_1-3c_2\Rightarrow 7c_1-3c_2=4~~~-(3)$

On solving equation 2 and 3, we get-

$c_1=1.3,c_2=1.7$

Puuting the above values in eqs.(1)-

Hence eqs.(1)$\implies$

$a_n=(1.3)7^n+(1.7)(-3)^n$

Given recurrence relation is-

$a_{n+2}=4a_{n+1}+21a_n$ Where, $a_o=3,a_1=4$

The characterstics equation of the above relation is given by-

$x^2=4x+21\\\Rightarrow x^2-4x-21=0\\ \Rightarrow (x-7)(x+3)=0$

Then the roots of equation are $7,-3$

So ,

$a_n=c_1(7)^n+c_2(-3)^n~~~~~-(1)$

At $n=0$

$a_o=c_1+c_1\Rightarrow c_1+c_2=3~~~~~-(2)$

At $n=1$

$a_1=7c_1-3c_2\Rightarrow 7c_1-3c_2=4~~~-(3)$

On solving equation 2 and 3, we get-

$c_1=1.3,c_2=1.7$

Puuting the above values in eqs.(1)-

Hence eqs.(1)$\implies$

$a_n=(1.3)7^n+(1.7)(-3)^n$

Given recurrence relation is-

$a_{n+2}=4a_{n+1}+21a_n$ Where, $a_o=3,a_1=4$

The characterstics equation of the above relation is given by-

$x^2=4x+21\\\Rightarrow x^2-4x-21=0\\ \Rightarrow (x-7)(x+3)=0$

Then the roots of equation are $7,-3$

So ,

$a_n=c_1(7)^n+c_2(-3)^n~~~~~-(1)$

At $n=0$

$a_o=c_1+c_1\Rightarrow c_1+c_2=3~~~~~-(2)$

At $n=1$

$a_1=7c_1-3c_2\Rightarrow 7c_1-3c_2=4~~~-(3)$

On solving equation 2 and 3, we get-

$c_1=1.3,c_2=1.7$

Puuting the above values in eqs.(1)-

Hence eqs.(1)$\implies$

$a_n=(1.3)7^n+(1.7)(-3)^n$

Given recurrence relation is-

$a_{n+2}=4a_{n+1}+21a_n$ Where, $a_o=3,a_1=4$

The characterstics equation of the above relation is given by-

$x^2=4x+21\\\Rightarrow x^2-4x-21=0\\ \Rightarrow (x-7)(x+3)=0$

Then the roots of equation are $7,-3$

So ,

$a_n=c_1(7)^n+c_2(-3)^n~~~~~-(1)$

At $n=0$

$a_o=c_1+c_1\Rightarrow c_1+c_2=3~~~~~-(2)$

At $n=1$

$a_1=7c_1-3c_2\Rightarrow 7c_1-3c_2=4~~~-(3)$

On solving equation 2 and 3, we get-

$c_1=1.3,c_2=1.7$

Puuting the above values in eqs.(1)-

Hence eqs.(1)$\implies$

$a_n=(1.3)7^n+(1.7)(-3)^n$