Solve the following recurrence relations.

b)

The given recurrence relation is

$$ \begin{aligned} x(n) &=3 x(n-1) \text { for } n>1, x(1)=4 \\ &=3[3 x(n-2)]=3^{2} x(n-2) \\ &=3\left(3[3 x(n-3)]=3^{3} x(n-3)\right.\\ \\ & \\ &=3^{i} x(n-i) \\ &=3^{n-1} x(n-1) \\ &=3^{n-1} x(1) \\ =& 4.3^{n-1} \end{aligned} $$

This can also be done using geometric progression

$$ \begin{aligned} x(n) &=x(1) \cdot q^{n-1} \\ &=4.3^{n-1} \end{aligned} $$

c)

The given recurrence relation is

\(x(n)=x(n-1)+n\) for \(n>0, x(0)=0\)

\(=[\mathrm{x}(\mathrm{n}-2)+(\mathrm{n}-1)]+\mathrm{n}=\mathrm{x}(\mathrm{n}-2)+(\mathrm{n}-1)+\mathrm{n}\)

\(=[x(n-3)+(n-2)+(n-1)]+n=x(n-3)+(n-2)+(n-1)+n\)

\(=x(n-i)+(n-i+1)+(n-i+2)+\ldots \ldots+n\)

\(=x(0)+1+2+3+\ldots .+n=\frac{n(n+1)}{2}\)

d)

The given recurrence relation is

$$ \begin{aligned} x(n) &=x(n / 2)+n \text { for } n>1, x(1)=1 \text {, we want to solve for } n=2^{k} \\ &=x\left(2^{k-1}\right)+2^{k} \\ &=\left[x\left(2^{k-2}\right)+2^{k-1}\right]+2^{k}=x\left(2^{k-2}\right)+2^{k-1}+2^{k} \\ &=\left[x\left(2^{k-3}\right)+2^{k-2}\right]+2^{k-1}+2^{k}=x\left(2^{k-3}\right)+2^{k-2}+2^{k-1}+2^{k} \\ \cdot \\ \cdot \\ \cdot \\ &=x\left(2^{k-k}\right)+2^{k-k+1}+2^{k-k+2}+\ldots .+2^{k} \\ =& x\left(2^{k-k}\right)+2^{1}+2^{2}+\ldots+2^{k} \\ =& 1+2^{1}+2^{2}+\ldots .+2^{k} \\ =& 2^{k+1}-1 \\ =& 2.2^{k}-1 \\ =& 2 n-1 \end{aligned} $$

e)

The given recurrence relation is

$$ \begin{aligned} \mathrm{x}(\mathrm{n}) &=\mathrm{x}(\mathrm{n} / 3)+1 \text { for } \mathrm{n}>1, \mathrm{x}(1)=1 \text { we want to solve for } n=3^{k} \\ &=x\left(3^{k}\right)+1 \\ &=\left[x\left(3^{k-2}\right)+1\right]+1=x\left(3^{k-2}\right)+2 \\ &=\left[x\left(3^{k-3}\right)+1\right]+2=x\left(3^{k-3}\right)+3 \end{aligned} $$

$$ =x\left(3^{k-i}\right)+i $$

$$ =x\left(3^{k-k}\right)+k=x(1)+k $$

Apply the logarithms on both sides for \(n=3^{k}\) then

$$ \begin{aligned} &=x(1)+k \\ &=1+\log _{3} n \end{aligned} $$