If $a_{n+1}=2a_n −n^2+n$ Define a sequence $a_n$ that satisfy the recurrence relation as described above, with $a_1 = 3$

Question

If $$a_{n+1}=2a_n −n^2+n$$

Define a sequence $a_n$ that satisfy the recurrence relation as described above, with $a_1 = 3$

Find the value of $$\dfrac{ |a_{20} - a_{15} | }{18133} $$

Attempt

First evaluate $a_{0}$

$$a_{1} = 2a_{0} \Rightarrow a_{0}= \frac{3}{2}$$

Then, use Z-transform: $$a_{n+1} - 2a_{n} + n^2 - n = 0$$ $$z(\mathbf{A}(z)-a_{0}) - 2\mathbf{A}(z) + \dfrac{z(z+1)}{(z-1)^3} - \dfrac{z}{(z-1)^2} = 0$$

$$\Rightarrow \mathbf{A}(z) = \dfrac{z(3z^3 -9z^2 + 9z - 7)}{2(z-2)(z-1)^3}$$ ​

$$\Rightarrow \mathbf{A}(z) = \dfrac{2z}{z-1} + \dfrac{z}{(z-1)^2} + \dfrac{z(z+1)}{(z-1)^3} - \dfrac{z}{2(z-2)}$$

The inverse of the Z-transform will be: $$\boxed{a_{n} = 2 + n + n^2 - 2^{n-1}}$$ ​

Now: $$a_{20} = 422-2^{19}$$ $$a_{15} = 242-2^{14}$$

Is it Correct??

Any other precise solution will be highly appreciated

Answer 1

As an alternative, you can use more elementary methods for linear difference equations. In fact, the general solution will be of the form $$ a_n = a_n^h + a_n^* $$

where $a_n^h$ is the general solution of the homogeneous equation, i.e. $a_n^h = c 2^n$, and $a_n^*$ is a particular solution of the full equation. If you try a particular solution "similar" to $-n^2+2n$, namely $a_n^* = k_1 n^2+ k_2n + k_3$, you'll see that $$ a_n = c 2^n +n^2+n+2 $$

The constant $c$ can be computed from the condition $a_1=3$, yielding the solution you have obtained using the Z-Transform.

Answer 2

Let $a_m=b_m+c_0+c_1m+c_2m^2+\cdots+c_rm^r$

$b_{m+1}+c_0+c_1(m+1)+c_2(m+1)^2+\cdots=2(b_n+c_0+c_1m+c_2n^2+\cdots)+n-n^2$

If $r\ge3,$ compare the coefficients of $m^r$

$$c_r=2c_r\iff c_r=0$$

$\implies a_m=b_m+c_0+c_1m+c_2m^2$

Compare the coefficients of $n^2, c_2=2c_1-1\iff c_2=1$

Similarly, comparing the coefficients of $n$ and the constants, $c_1=1, c_0=2$

so that $b_{m+1}=2b_m=\cdots=2^tb_{m-t}$ for integer $t\ge0$

Again $3=a_1=b_1+c_0+c_1+c_2\implies b_1=-1$