Question

Use generating functions to solve the recurrence relation $a_{\mathrm{k}}=7a_{\mathrm{k}-1}$with the initial condition$a_0=5$.

Short answer

$a_k=5·7^k$

Step-by-step solution

Use Generating Function:

Generating function for the sequence $a_0,a_1,\dots ,a_k,\dots$of real numbers is the infinite series:

$$\begin{gathered} G\left(x\right)=a_0+a_{1}x+a_2{x}^2+\dots +a_k{x}^k+\dots \\ =\sum _{k=0}^{+\infty }{a}_k{x}^k \end{gathered}$$

Extended binomial theorem:

localid="1668588158853" ${(1+x)}^u=\sum _{k=0}^{+\infty }\left(\begin{array}{c}u\\ k\end{array}\right)x^{\mathrm{k}}$

Let

$$\begin{gathered} G\left(x\right)-a_0=\sum _{k=1}^{+\infty }{a}_k{x}^k \\ =\sum _{k=1}^{+\infty }7a_{k-1}{x}^k{a}_k=7a_{k-1}Whenk\ge 1 \\ =7x\sum _{k=1}^{+\infty }{a}_{k-1}{x}^{k-1} \\ =7x\sum _{m=0}^{+\infty }{a}_m{x}^{m}Letm=k-1 \\ =7xG\left(x\right) \end{gathered}$$

Calculate the expression forG(x).

$$\begin{gathered} G\left(x\right)-a_0=7xG\left(x\right)G\left(x\right)-5 \\ =7xG\left(x\right)G\left(x\right)-7xG\left(x\right)-5 \\ =0(1-7x)G\left(x\right)-5 \\ =0(1-7x)G\left(x\right) \\ =5G\left(x\right) \\ =\frac{5}{1-7x} \end{gathered}$$

Now term the sequence $\left\{a_k\right\}$

$$\begin{gathered} G\left(x\right)=\frac{5}{1-7x} \\ =5{(1+(-7x\left)\right)}^{-1} \\ =5\sum _{k=0}^{+\infty }\left(\begin{array}{c}-1\\ k\\ \end{array}\right){(-7x)}^k \\ =5\sum _{k=0}^{+\infty }\left(\begin{array}{c}-1\\ k\\ \end{array}\right){(-1)}^k{7}^k{x}^k \\ =5\sum _{k=0}^{+\infty }{7}^k{x}^k \\ =\sum _{k=0}^{+\infty }5·7^k{x}^k \end{gathered}$$

Thus,$a_k=5·7^k$