Answer:
a) [tex]a_n=(-1)^n \cdot 5[/tex]
b)[tex]a_n = 1 + 3n[/tex]
c) [tex]a_n=4-\frac{n(n+1)}{2}[/tex]
d) [tex]a_n=3 \cdot 2^{\frac{n(n+1)}{2}}[/tex]
Step-by-step explanation:
We solve this using backward or forward substitution.
a) We have this:
[tex]a_1= - a_0= (-1)a_0[/tex]
then:
[tex] a_2= -a_1= (-1)(-1)a_0=(-1)^2 a_0[/tex]
for [tex] n=3[/tex] we have:
[tex] a_3= -a_2= (-1)(-1)^2 a_0=(-1)^3 a_0[/tex]
from this, we can see that [tex] a_n = (-1)^n a_0[/tex] is a solution for this recurrence relation, where [tex] a_0=5[/tex]. This is:
[tex] a_n=(-1)^n\cdot 5[/tex]
b) We have [tex]a_n=a_{n-1}+3[/tex] with [tex] a_0=1 [/tex]. Then:
[tex] a_1=a_0+3=1+3=4\\
a_2=a_1+3=4+3=7[/tex] but at the same time
[tex] a_2 = a_1 + 3 =(a_0+3)+3 = a_0+ 2 \cdot 3 [/text]
for [tex]a_3[/tex] we have:
[tex] a_3=a_2+3=7+10=4[/tex] or [tex]a_3=a_2+3=(a_0+2\cdot 3)+3=a_0+3\cdot 3[/tex]
by the next:
[tex]a_4 = a_3 + 3 = (a_0 + 3\cdot 3)+3 = a_0 + 4\cdot 3[/tex]
We can see that the recurrence rule is:
[tex]a_n=a_0+n\cdot 3[/tex]
this is [tex]a_n=1+n\cdot 3[/tex]
c)Note that:
[tex]a_1-a_0 = (a_0 - 1)-a_0=-1\\a_2-a_1 = (a_1 - 2) -a_1 = -2\\a_3-a_2 = (a_2 - 3) - a_2 = -3\\\ldots\\a_n-a_{n-1} = (a_{n-1}-n)-a_{n-1} =-n[/tex]
taking all this we have to:
[tex]a_n-a_0=\sum\limits_{k=1}^n (a_k - a_{k-1}) =\sum\limits_{k=1}^n -k = -\sum\limits_{k=1}^n k = - \frac{n(n+1)}{2}[/tex]
then:
[tex]a_n=a_0-\frac{n(n+1)}{2}[/tex]
this is:
[tex]a_n=4-\frac{n(n+1)}{2}[/tex]
d)We take [tex]a_n=2^na_{n-1}[/tex]. Then:
[tex]a_n=2^na_{n-1}=2^n(2^{n-1}a_{n-2}) = 2^n\cdot 2^{n-1}(2^{n-2}a_{n-3}) = \dots =2^n\cdot2^{n-1}\cdot2^{n-2}\cdots 2^1 a_0=2^{n+(n-1)+(n-2)+\dots + 1}a_0=2^{\frac{n(n+1)}{2}}a_0[/tex]
replacing [tex]a_0=3[/tex]we have:
[tex]a_n=3 \cdot 2^{\frac{n(n+1)}{2}}[/tex]
