gainejavima

2021-11-19

Consider the following linear difference equation

with

Befory

Beginner2021-11-20Added 19 answers

Apart from trying what Will Jagy said, you can also note that, for $n=2,3,4,\dots ,$

$\sum _{k=2}^{n}(n-k+1)\cdot (-1)=\sum _{k=2}^{n}(n-k+1)({f}_{k}-2{f}_{k-1}+{f}_{k-2})$

$={f}_{n}+\sum _{k=2}^{n-1}((n-k+{1}_{-2}(n-k)+(n-k-1)){f}_{k}+((n-2)-2(n-1))f\left(1\right)+(n-1)f\left(0\right)$

$={f}_{n}-nf\left(1\right)+(n-1)f\left(0\right)$

Thus, for every$n=0,1,2,\dots ,$

$f}_{n}\left(n\right)=nf\left(1\right)-(n-1)f\left(0\right)-\sum _{k=2}^{n}(n-k+1)=nf\left(1\right)-(n-1)f\left(0\right)-\frac{n(n-1)}{2$

Thus, for every

James Kilian

Beginner2021-11-21Added 20 answers

Easily you can determine that for the homogeneous recurrence equation

${f}_{k+1}^{h}-{2}_{k}^{h}+{f}_{k-1}^{h}=0$

the solution is

${f}_{k}^{h}={c}_{1}+{c}_{2}k$

now, a particular solution which should be polynomial, you can propose

$f}_{k}^{p}={c}_{1}+{c}_{2}k+{c}_{3}{k}^{2$

and the coefficients

the solution is

now, a particular solution which should be polynomial, you can propose

and the coefficients

22+64

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