Leaving cert maths higher level REcurrance equations need help

Liamo_mu

how to i do the follwing
n n
Un=1/2[4 -2 ]

n
Un+1-2Un-4 =0


the n should be over the 4 and 2 on first part

the n should be over the 4 on the second part

Michael Collins

So you're trying to show [latex] u_{n}[/latex], given by

[latex] \displaystyle u_{n}=\frac{1}{2}(4^{n}-2^{n}) [/latex]

satisfies the difference equation

[latex] \displaystyle u_{n+1}- 2u_{n}-4^{n}=0[/latex].

You can simplify [latex] u_{n}[/latex] using the rules of indices

[latex] \displaystyle u_{n}=\frac{1}{2}(4^{n}-2^{n})=2^{-1}\left((2^{2})^{n}-2^{n}\right) = 2^{-1}\left((2^{2n}-2^{n}\right)=2^{2n-1}-2^{n-1}[/latex]

Then

[latex] \displaystyle u_{n+1} =2^{2(n+1)-1}-2^{(n+1)-1} [/latex]

i.e. replace every "n" by "(n+1)". This simplifies to

[latex] \displaystyle u_{n+1} =2^{2n+1}-2^{n} [/latex].

Now take away two times [latex]u_{n}[/latex] and what do you get? Finally take away [latex]4^{n}[/latex], you should get zero, i.e. the right hand side of the difference equation above.