• Support PF! Buy your school textbooks, materials and every day products Here!

Show that this sequence satisfies the recurrence relation

  • Thread starter bdh2991
  • Start date
  • #1
103
0

Homework Statement


Let d0, d1, d2,... be defined by the formula dn = 3n - 2n for all integers n ≥ 0. Show that this sequence satisfies the recurrence relation.

dk = 5dk-1 - 6dk-2.


Homework Equations





The Attempt at a Solution



I found that dk = 3k - 2k

dk-1 = 3k-1 - 2k-1
dk-2 = 3k-2 - 2k-2

after plugging dk-1 and dk - 2 into the formula dk = 5dk-1 - 6dk-2, i am stuck and do not understand how to do the algebra if that is what i 'm supposed to be doing...any help?
 

Answers and Replies

  • #2
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,313
1,003

Homework Statement


Let d0, d1, d2,... be defined by the formula dn = 3n - 2n for all integers n ≥ 0. Show that this sequence satisfies the recurrence relation.

dk = 5dk-1 - 6dk-2.

Homework Equations



The Attempt at a Solution



I found that dk = 3k - 2k

dk-1 = 3k-1 - 2k-1
dk-2 = 3k-2 - 2k-2

after plugging dk-1 and dk - 2 into the formula dk = 5dk-1 - 6dk-2, i am stuck and do not understand how to do the algebra if that is what i 'm supposed to be doing...any help?
Take
dk-1 = 3k-1 - 2k-1

and

dk-2 = 3k-2 - 2k-2 '​
Plug those into
5dk-1 - 6dk-2 .​
Do some algebra & see what you get.
 
  • #3
103
0
Take
dk-1 = 3k-1 - 2k-1

and

dk-2 = 3k-2 - 2k-2 '​
Plug those into
5dk-1 - 6dk-2 .​
Do some algebra & see what you get.
This is what i tried doing at i stated above...after you plug them in i get

3k - 2k = 5( 3k-1 - 2k-1) - 6(3k-2 - 2k-2), I don't really understand what i can do with that algebraically...i must just be missing it...
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
This is what i tried doing at i stated above...after you plug them in i get

3k - 2k = 5( 3k-1 - 2k-1) - 6(3k-2 - 2k-2), I don't really understand what i can do with that algebraically...i must just be missing it...
Use 3^(k-1)=3*3^(k-2) and 2^(k-1)=2*2^(k-2).
 
  • #5
103
0
Use 3^(k-1)=3*3^(k-2) and 2^(k-1)=2*2^(k-2).
ok so changing the equation to that gives me:

5(3*3k-2 - 2*2k-2) - 6( 3k-2 - 2k-2)

i can see how that gave me some like terms but multiplying through gives me:

15 * 3k-2 - 10 * 2k-2 - 6 * 3k-2 - 6 * 2k-2

i feel like that wasn't where you were leading me lol...i'm sorry, i'm not that great at algebra
 
  • #6
Dick
Science Advisor
Homework Helper
26,258
618
ok so changing the equation to that gives me:

5(3*3k-2 - 2*2k-2) - 6( 3k-2 - 2k-2)

i can see how that gave me some like terms but multiplying through gives me:

15 * 3k-2 - 10 * 2k-2 - 6 * 3k-2 - 6 * 2k-2

i feel like that wasn't where you were leading me lol...i'm sorry, i'm not that great at algebra
I guess not. But you're almost there. Collect the 3^(k-2) terms. What do you get? And you've a sign error on the last term. Could you fix it?
 

Related Threads on Show that this sequence satisfies the recurrence relation

Replies
6
Views
1K
Replies
5
Views
466
Replies
9
Views
912
Replies
1
Views
662
Replies
11
Views
5K
Replies
1
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
9
Views
549
Top