Finding a formula for a sequence

Question

Let $a_n$ be a sequence such that $a_0=5$ and $a_{n}+a_{n+1}=3$ for all $n$ greater than $0$.

I defined a function $f(x)$ whose coefficients are same of that mentioned sequence. I was able to get $$f(x)=\displaystyle \sum_{n=0}^\infty [ 5 (-1)^n x^n + 3 (-1)^n x^{n+1} + 3 x^{2n+1} ] $$ But couldn't make it far. How do you find a formula for $a_n$ given that $a_1$ is just $1$?

Answer 1

Define $b_n=a_n-1.5$. Therefore$$b_n+b_{n+1}=0$$which means that$$b_{n+1}=b_1(-1)^n$$or $$a_n=1.5+k(-1)^n$$with $a_1=1$ we have $$a_1=1=1.5+k(-1)\to k= 0.5$$therefore $$a_n={3+(-1)^n\over 2}$$

Answer 2

Let an be a sequence such that $a_0=5$ and $a_n+a_{n+1}=3$ for all $n$ greater than $0$.

First note: $a_0+a_1=3 \Rightarrow 5+a_1=3 \Rightarrow a_1=-2$.

It looks you are trying to use the generating function $f(x)=\sum_{n=0}^{\infty} a_nx^n$. Here are the steps: $$\sum_{n=0}^{\infty} a_nx^{n+1}+\sum_{n=0}^{\infty} a_{n+1}x^{n+1}=3\sum_{n=0}^{\infty} x^{n} \Rightarrow \\ xf(x)+f(x)-a_0=3\cdot \frac1{1-x} \Rightarrow \\ f(x)=\frac{5-2x}{(1-x)(1+x)}=\frac7{2(1+x)}+\frac3{2(1-x)}=\\ \frac72\sum_{n=0}^{\infty}(-x)^n+\frac32\sum_{n=0}^{\infty} x^n =\\ \sum_{n=0}^{\infty}\left[\frac72(-1)^n+\frac32\right]x^n \Rightarrow \\ a_n=\frac72(-1)^n+\frac32.$$

Can you solve the recurrence relation $a_n+a_{n+1}=3$, if $a_0=2, a_1=1$?

Answer 3

Another way is to use characteristic polynomial method (here is another example). Given $$a_n+a_{n+1}=3$$ $$a_{n-1}+a_n=3$$ we have $a_{n+1}-a_{n-1}=0$, with characteristic polynomial $$x^2-1=0$$ with solutions $-1,1$ and general term $$a_n=A\cdot (-1)^n+B\cdot 1^n=A\cdot (-1)^n+B \tag{1}$$ From $$a_0=5=A+B$$ $$a_1=-2=-A+B$$ we have $B=\frac{3}{2}$ and $A=\frac{7}{2}$ or $$a_n=\frac{7}{2}\cdot (-1)^n+\frac{3}{2} \tag{2}$$