# Discrete mathematics If x_{1}=2, x_{n}=4X_{n-1}-4n \forall n \geq 2. Find t

Discrete mathematics
If ${x}_{1}=2,{x}_{n}=4{X}_{n-1}-4n\mathrm{\forall }n\ge 2$.
Find the general term xn.
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Step 1
Given recurrence relation is
${x}_{n}=4{x}_{n-1}-4n$, ${x}_{1}=2$
Rewriting given recurrence relation, we have
${x}_{n}-4{x}_{n-1}=-4n$
Associated homogeneous recurrence relation is
${x}_{n}-4{x}_{n-1}=0$
Auxiliary equation is
$m-4=0$
$⇒m=4$
$⇒$ Complementary solution is
${x}_{n}^{c}=C{\left(4\right)}^{n}$ C is arbitrary constant.
Step 2
To find particular solution:
the non-homogeneous function is $Q\left(n\right)=-4n=0{\left(1\right)}^{n}-4n{\left(1\right)}^{n}$
$⇒$ particular solution will have form
${x}_{n}^{p}=A+Bn$
$⇒{x}_{n-1}^{p}=A+B\left(n-1\right)=A+Bn-B$
Substituting in given recurrence relation
$A+Bn=4\left(A+Bn-B\right)-4n$
$⇒A+Bn-4A-4Bn+4B=-4n$
$⇒\left(-3A+4B\right)+\left(-3B\right)n=-4n$
$⇒-3A+4B=0,-3B=-4$ by comparing coefficients of 1&n.
$⇒B=\frac{4}{3}$ and $3A=4B=4×\frac{4}{3}$
$⇒A=\frac{16}{9}$ and $B=\frac{4}{3}$
Hence particular solution is
${x}_{n}^{p}=A+Bn=\frac{16}{9}+\frac{4}{3}n$
$⇒{x}_{n}^{p}=\frac{16}{9}+\frac{4}{3}n$
Step 3
General solution is
${x}_{n}={x}_{n}^{c}+{x}_{n}^{p}$
${x}_{n}=C{\left(4\right)}^{n}+\frac{16}{9}+\frac{4}{3}n$
Given that, ${x}_{1}=2$
$⇒2=C{\left(4\right)}^{1}+\frac{16}{9}+\frac{4}{3}×1$
$⇒2=4C+\frac{16+12}{9}$