Question

Let${\mathrm{a}}_{\mathrm{n}}=2^{\mathrm{n}}+5\cdot 3^{\mathrm{n}}\text{for}\mathrm{n}=0,1,2,\dots$

a) Find ${\mathrm{a}}_0,{\mathrm{a}}_1,{\mathrm{a}}_2,{\mathrm{a}}_3\text{and}{\mathrm{a}}_4$

b) Show that ${\mathrm{a}}_2=5{\mathrm{a}}_1-6{\mathrm{a}}_0,{\mathrm{a}}_3=5{\mathrm{a}}_2-6{\mathrm{a}}_1\text{and}{\mathrm{a}}_4=5{\mathrm{a}}_3-6{\mathrm{a}}_2$

c) Show that${\mathbf{a}}_{\mathbf{n}}\mathbf{=}\mathbf{5}{\mathbf{a}}_{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{-}\mathbf{6}{\mathbf{a}}_{\mathbf{n}\mathbf{-}\mathbf{2}}\mathbf{\text{for all integers}}\mathbf{n}\mathbf{\text{with}}\mathbf{n}\mathbf{\ge }\mathbf{2}$

Short answer

$\begin{array}{r}6,17,49,143,421\\ {\mathrm{a}}_2=5{\mathrm{a}}_1-6{\mathrm{a}}_0,{\mathrm{a}}_3=5{\mathrm{a}}_2-6{\mathrm{a}}_1,{\mathrm{a}}_4=5{\mathrm{a}}_3-6{\mathrm{a}}_2\\ {\mathrm{a}}_0=5{\mathrm{a}}_{\mathrm{n}-1}-6{\mathrm{a}}_{\mathrm{n}-2},\mathrm{n}\ge 2\end{array}$

Step-by-step solution

Step 1

A recurrence relation for the sequence$\left\{{\mathrm{a}}_{\mathrm{n}}\right\}$ is an equation that expressesin terms of one or more of the previous terms of the sequence, namely,${\mathrm{a}}_0,{\mathrm{a}}_1,........{\mathrm{a}}_{\mathrm{n}-1}$ , for all integers n with$\mathrm{n}>{\mathrm{n}}_0$ , where${\mathrm{n}}_0$ is a nonnegative integer. A sequence is called a solution of a recurrence relation if its terms satisfy the recurrence relation..

Step 2

a) Given

$\begin{array}{r}{\mathrm{a}}_{\mathrm{n}}=2^{\mathrm{n}}+5\cdot 3^{\mathrm{n}}\\ \mathrm{n}=0,1,2,\dots \end{array}$

Replace in the given expression for ${\mathrm{a}}_{\mathrm{n}}$with 0,1,2,3,4 and then evaluate:

$\begin{array}{r}{\mathrm{a}}_0=2^0+5\cdot 3^0=1+5=6\\ {\mathrm{a}}_1=2^1+5\cdot 3^1=2+15=17\\ {\mathrm{a}}_2=2^2+5\cdot 3^2=4+45=49\\ {\mathrm{a}}_3=2^3+5\cdot 3^3=8+135=143\\ {\mathrm{a}}_4=2^4+5\cdot 3^4=16+405=421\end{array}$

Step 3

Let us determine $5{\mathrm{a}}_1-6{\mathrm{a}}_0$

$5{\mathrm{a}}_1-6{\mathrm{a}}_0=5\left(17\right)-6\left(6\right)=85-36=49={\mathrm{a}}_2$

Let us determine $5_{\mathrm{a}}-6{\mathrm{a}}_1$

$5{\mathrm{a}}_2-6{\mathrm{a}}_1=5\left(49\right)-6\left(17\right)=245-102=143={\mathrm{a}}_3$
Let us determine $5_{\mathrm{a}}-6{\mathrm{a}}_1$

$5{\mathrm{a}}_3-6{\mathrm{a}}_3=5\left(143\right)-6\left(49\right)=715-294=421={\mathrm{a}}_4$

Step 4

c) given

${\mathrm{a}}_{\mathrm{n}}=2^{\mathrm{n}}+=5.3^{\mathrm{n}}$

To prove: ${\mathrm{a}}_{\mathrm{n}}=5{\mathrm{a}}_{\mathrm{n}-1}-6{\mathrm{a}}_{\mathrm{n}-2}\mathrm{n}\ge 2$

Replace n in ${\mathrm{a}}_{\mathrm{n}}=2^{\mathrm{n}}+5.3^{\mathrm{n}}$by n - 1 :

${\mathrm{a}}_{\mathrm{n}-1}=2^{\mathrm{n}-1}+5\cdot 3^{\mathrm{n}-1}$

Replace n in${\mathrm{a}}_{\mathrm{n}}=2^{\mathrm{n}}+5\cdot 3^{\mathrm{n}-1}$by n - 2 :

${\mathrm{a}}_{\mathrm{n}-2}=2^{\mathrm{n}-2}+5.3^{\mathrm{n}-2}$

We will start from the expression $a_n=5a_{n-1}-6a_{n-2}$and prove that this term has to be equal to $a_n$(when > 2). Let us use the two previous expressions derived for and $a_{n-2}$

$\begin{array}{r}5{\mathrm{a}}_{\mathrm{n}-1}-6{\mathrm{a}}_{\mathrm{n}-2}=5\left(2^{\mathrm{n}-1}+5\cdot 3^{\mathrm{n}-1}\right)-6\left(2^{\mathrm{n}-1}+5\cdot 3^{\mathrm{n}-1}\right)\\ =5\cdot 2^{\mathrm{n}-1}+5\cdot 5\cdot 3^{\mathrm{n}-1}-6\cdot 2^{\mathrm{n}-2}-6\cdot 5\cdot 3^{\mathrm{n}-2}\\ \mathrm{Let}\mathrm{us}\mathrm{group}\mathrm{the}\mathrm{terms}\mathrm{contains}\mathrm{powers}\mathrm{of}:\\ =\left(5\cdot 2^{\mathrm{n}-1}-6\cdot 2^{\mathrm{n}-2}\right)+\left(5\cdot 5\cdot 3^{\mathrm{n}-1}-6\cdot 5\cdot 3^{\mathrm{n}-2}\right)\\ =\left(5\cdot 2\cdot 2^{\mathrm{n}-1}-6\cdot 2^{\mathrm{n}-2}\right)+\left(5\cdot 5\cdot 3\cdot 3^{\mathrm{n}-1}-6\cdot 5\cdot 3^{\mathrm{n}-2}\right)\end{array}$

Let us factor out $2^{n-2}$from the first term and $3^{n-2}$from the second term

$\begin{array}{r}=2^{\mathrm{n}-2}(5\cdot 2-6)+3^{\mathrm{n}-2}(5\cdot 5\cdot 3-6\cdot 5)\\ =2^{\mathrm{n}-2}(10-6)+3^{\mathrm{n}-2}(75-30)\\ =2^{\mathrm{n}-2}\left(4\right)+3^{\mathrm{n}-2}\left(45\right)\\ =2^{\mathrm{n}-2}\left(2^2\right)+3^{\mathrm{n}-2}\left(5\cdot 3^2\right)\\ =2^{\mathrm{n}}+5\cdot 3^{\mathrm{n}}\\ ={\mathrm{a}}_{\mathrm{n}}\end{array}$

Hence, proved