the question is

''Find the solution to the equations.

(i) $\displaystyle 25T_{n+2}=T_{n}\\T_{0}=125\\T_{1}=25$

(ii) $\displaystyle 25T_{n+2}=-T_{n}\\T_{0}=125\\T_{1}=25$

for i) i have got

If $\displaystyle T_{n}=x^n$, so $\displaystyle 25x^{n+2}-x^n=0\\x^n(25x^2-1)=0\\x=0.2$

general solution: $\displaystyle T_{n}=A(1/5)^n\\T_{0}=A=125\\A=125$

therefore actual solution is $\displaystyle T_{n}=125*(1/5)^n$

for ii) i have got

If

$\displaystyle T_{n}=x^n$,

so

$\displaystyle 25x^{n+2}+x^n=0\\x^n(25x^2+1)=0\\x=\pm 0.2i$

general solution:

$\displaystyle T_{n}=A\cos(0.2n)+B\sin(0.2n)$

where A,B are constants.

after using

$\displaystyle T_{0}=125$,

I've got A=125

I'm wondering is my general solution incorrect and how can I find the constant B?