Find the first six terms of the sequence defined by each of these recurrence relations and initial conditions.
a) an=6an-1, a0=2
b) an=a2n-1, a1=2
c) an=an-1+3an-2, a0=1 , a1=2
d) an=nan-1+n2 an-2 a0=1 ,a1=1
e) an= an-1+an-3 , a0=1, a1=2 , a2=0

Answers

Answer 1

Answer:

See explanation

Step-by-step explanation:

Solving (a):

[tex]a_n = 6a_{n-1}[/tex] where [tex]a_0 = 2[/tex]

n = 1

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_1 = 6a_{1-1}[/tex]

[tex]a_1 = 6a_{0}[/tex]

Substitute 2 for [tex]a_0[/tex]

[tex]a_1= 6 * 2[/tex]

[tex]a_1= 12[/tex]

n = 2

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_2 = 6a_{2-1}[/tex]

[tex]a_2 = 6a_{1}[/tex]

Substitute 12 for [tex]a_1[/tex]

[tex]a_2= 6 * 12[/tex]

[tex]a_2= 72[/tex]

n = 3

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_3 = 6a_{3-1}[/tex]

[tex]a_3 = 6a_2[/tex]

Substitute 72 for [tex]a_2[/tex]

[tex]a_3= 6 * 72[/tex]

[tex]a_3= 432[/tex]

n = 4

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_4 = 6a_{4-1}[/tex]

[tex]a_4 = 6a_{3}[/tex]

Substitute 432 for [tex]a_3[/tex]

[tex]a_4 = 6 * 432[/tex]

[tex]a_4 = 2592[/tex]

n = 5

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_5 = 6a_{5-1}[/tex]

[tex]a_5 = 6a_4[/tex]

Substitute 2592 for [tex]a_4[/tex]

[tex]a_5 = 6 * 2592[/tex]

[tex]a_5 = 15552[/tex]

n = 6

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_6 = 6a_{6-1}[/tex]

[tex]a_6 = 6a_{5}[/tex]

Substitute 15552 for [tex]a_5[/tex]

[tex]a_6 = 6 * 15552[/tex]

[tex]a_6 = 93312[/tex]

[tex]a_1= 12[/tex]   [tex]a_2= 72[/tex]   [tex]a_3= 432[/tex]   [tex]a_4 = 2592[/tex]   [tex]a_5 = 15552[/tex]   [tex]a_6 = 93312[/tex]

Solving (b):

[tex]a_n = a^2_{n - 1[/tex] where [tex]a_1 = 2[/tex]

We have

[tex]a_1 = 2[/tex] which serves as the first term

n =2

[tex]a_n = a^2_{n - 1[/tex]

[tex]a_2 = a^2_{2-1}[/tex]

[tex]a_2 = a^2_{1}[/tex]

Substitute 2 for [tex]a_1[/tex]

[tex]a_2 = 2^2[/tex]

[tex]a_2 = 4[/tex]

n = 3

[tex]a_3 = a^2_{3-1}[/tex]

[tex]a_3 = a^2_{2}[/tex]

[tex]a_3 = 4^2[/tex]

[tex]a_3 = 16[/tex]

n = 4

[tex]a_4 = a^2_{4-1}[/tex]

[tex]a_4 = a^2_3[/tex]

[tex]a_4 = 16^2[/tex]

[tex]a_4 = 256[/tex]

n =5

[tex]a_5 = a^2_{5-1[/tex]

[tex]a_5 = a^2_4[/tex]

[tex]a_5 = 256^2[/tex]

[tex]a_5 = 65536[/tex]

n = 6

[tex]a_6 = a^2_{6-1[/tex]

[tex]a_6 = a^2_{5[/tex]

[tex]a_6 = 65536^2[/tex]

[tex]a_6 = 4294967296[/tex]

[tex]a_1 = 2[/tex]   [tex]a_2 = 4[/tex]   [tex]a_3 = 16[/tex]   [tex]a_4 = 256[/tex]   [tex]a_5 = 65536[/tex]   [tex]a_6 = 4294967296[/tex]

Solving (c):

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] [tex]a_0=1[/tex]  ; ;  [tex]a_1=2[/tex]

[tex]a_1=2[/tex] ---- First term

n = 2

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_2=a_{2-1}+3a_{2-2}[/tex]

[tex]a_2=a_1+3a_0[/tex]

Substitute values for a1 and a0

[tex]a_2=2+3 * 1[/tex]

[tex]a_2=2+3[/tex]

[tex]a_2=5[/tex]

n = 3

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_3=a_{3-1}+3a_{3-2}[/tex]

[tex]a_3=a_{2}+3a_{1}[/tex]

[tex]a_3=5+3 * 2[/tex]

[tex]a_3=5+6[/tex]

[tex]a_3=11[/tex]

n = 4

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_4=a_{4-1}+3a_{4-2}[/tex]

[tex]a_4=a_{3}+3a_{2}[/tex]

[tex]a_4=11+3 * 5[/tex]

[tex]a_4=11+15[/tex]

[tex]a_4=26[/tex]

n = 5

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_5=a_{5-1}+3a_{5-2};[/tex]

[tex]a_5=a_{4}+3a_3[/tex]

[tex]a_5=26+3 * 11[/tex]

[tex]a_5=26+33[/tex]

[tex]a_5=59[/tex]

n = 6

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_6=a_{6-1}+3a_{6-2}[/tex]

[tex]a_6=a_{5}+3a_4[/tex]

[tex]a_6=59+3*26[/tex]

[tex]a_6=59+78[/tex]

[tex]a_6=137[/tex]

[tex]a_1=2[/tex]   [tex]a_2=5[/tex]   [tex]a_3=11[/tex]   [tex]a_4=26[/tex]   [tex]a_5=59[/tex]   [tex]a_6=137[/tex]

Solving (d):

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex];     [tex]a_0=1[/tex]; [tex]a_1=1[/tex]

[tex]a_1=1[/tex] --- First term

n = 2

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_2=2 * a_{2-1}+2^2a_{2-2}[/tex]

[tex]a_2=2 * a_1+4*a_0[/tex]

[tex]a_2=2 * 1+4*1[/tex]

[tex]a_2=2 +4[/tex]

[tex]a_2=6[/tex]

n = 3

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_3=3 * a_{3-1}+3^2 * a_{3-2}[/tex]

[tex]a_3=3 * a_{2}+9 * a_{1}[/tex]

[tex]a_3=3 * 6+9 * 1[/tex]

[tex]a_3=18+9[/tex]

[tex]a_3=27[/tex]

n = 4

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_4=4*a_{4-1}+4^2*a_{4-2}[/tex]

[tex]a_4=4*a_{3}+16*a_{2}[/tex]

[tex]a_4=4*27+16*6[/tex]

[tex]a_4=204[/tex]

n = 5

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_5=5 * a_{5-1}+5^2 * a_{5-2}[/tex]

[tex]a_5=5 * a_{4}+25 * a_{3}[/tex]

[tex]a_5=5 * 204+25 *27[/tex]

[tex]a_5=1695[/tex]

n = 6

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_6=6 * a_{6-1}+6^2*a_{6-2}[/tex]

[tex]a_6=6 * a_{5}+36*a_{4}[/tex]

[tex]a_6=6 * 1695+36*204[/tex]

[tex]a_6=17514[/tex]

[tex]a_1=1[/tex]   [tex]a_2=6[/tex]   [tex]a_3=27[/tex]   [tex]a_4=204[/tex]   [tex]a_5=1695[/tex]   [tex]a_6=17514[/tex]

Solving (e):

[tex]a_n= a_{n-1}+a_{n-3};\ a_0=1; a_1=2; a_2=0[/tex]

First term: [tex]a_1=2[/tex]

Second Term: [tex]a_2=0[/tex]

n = 3

[tex]a_n= a_{n-1}+a_{n-3}[/tex] becomes

[tex]a_3= a_{3-1}+a_{3-3}[/tex]

[tex]a_3= a_{2}+a_0[/tex]

[tex]a_3= 0+1[/tex]

[tex]a_3= 1[/tex]

n = 4

[tex]a_n= a_{n-1}+a_{n-3}[/tex] becomes

[tex]a_4= a_{4-1}+a_{4-3}[/tex]

[tex]a_4= a_{3}+a_{1}[/tex]

[tex]a_4= 1+2[/tex]

[tex]a_4=3[/tex]

n = 5

[tex]a_n= a_{n-1}+a_{n-3}[/tex] becomes

[tex]a_5= a_{5-1}+a_{5-3}[/tex]

[tex]a_5= a_{4}+a_{2}[/tex]

[tex]a_5= 3 + 0[/tex]

[tex]a_5= 3[/tex]

n = 6

[tex]a_n= a_{n-1}+a_{n-3}[/tex] becomes

[tex]a_6= a_{6-1}+a_{6-3}[/tex]

[tex]a_6= a_{5}+a_{3}[/tex]

[tex]a_6= 3 + 1[/tex]

[tex]a_6= 4[/tex]

[tex]a_1=2[/tex]   [tex]a_2=0[/tex]   [tex]a_3= 1[/tex]   [tex]a_4=3[/tex]   [tex]a_5= 3[/tex]   [tex]a_6= 4[/tex]


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Answers

Answer:  

               「0.8   0   -2

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Step-by-step explanation:

The [tex]3\times 3[/tex] matrix will be "[tex]\left[\begin{array}{ccc}0.3&0&-1.8\\0&1.3&3.9\\0&0&1\end{array}\right][/tex]".

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→ [tex]A = ST[/tex]

      [tex]= \left[\begin{array}{ccc}0.3&0&0\\0&1.3&0\\0&0&1\end{array}\right] \ \left[\begin{array}{ccc}1&0&-6\\0&1&3\\0&0&1\end{array}\right][/tex]

      [tex]= \left[\begin{array}{ccc}0.3&0&-0.3\times 6\\0&1.3&1.3\times 3\\0&0&1\times 1\end{array}\right][/tex]

      [tex]= \left[\begin{array}{ccc}0.3&0&-1.8\\0&1.3&3.9\\0&0&1\end{array}\right][/tex]

Thus the above answer is correct.

       

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Step-by-step explanation:

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