Let bo, b,, bz, ... be the sequence defined by the explicit formula b, = C- 3" + D(-2)^ for every integer n2 0, where C and D are real numbers. Fill in the blanks in the following proof to show that for any choice of C and D and for each integer k 2 2, b = bk -1+ 6bk – 2° Proof: Let C and D be any real numbers and let k be any integer such that k 2 2. By definition of bo, b,, b,, ..., bx -1 = and bk - 2 = To complete the proof, you need to show that b = b -1+ 6bk - 2. When the expressions for b-1 and b- 2 are substituted into b -1+ 6b - 2, the result is | -«([ bk - 1+ 6b 3k-1 + 6(- Further simplification gives that
Let bo, b,, bz, ... be the sequence defined by the explicit formula b, = C- 3" + D(-2)^ for every integer n2 0, where C and D are real numbers. Fill in the blanks in the following proof to show that for any choice of C and D and for each integer k 2 2, b = bk -1+ 6bk – 2° Proof: Let C and D be any real numbers and let k be any integer such that k 2 2. By definition of bo, b,, b,, ..., bx -1 = and bk - 2 = To complete the proof, you need to show that b = b -1+ 6bk - 2. When the expressions for b-1 and b- 2 are substituted into b -1+ 6b - 2, the result is | -«([ bk - 1+ 6b 3k-1 + 6(- Further simplification gives that
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Let b0, b1, b2, be the sequence defined by the explicit formula bn = C · 3n + D(−2)n for every integer n ≥ 0,
where C and D are real numbers. Fill in the blanks in the following proof to show that for any choice of C and D and for each integer k ≥ 2,
bk = bk − 1 + 6bk − 2.
Proof: Let C and D be any real numbers and let k be any integer such that
k ≥ 2.
By definition of b0, b1, b2, ,![Let bo, b,, b2, ... be the sequence defined by the explicit formula
b. = C. 3" + D(-2)" for every integer n 2 0,
where C and D are real numbers. Fill in the blanks in the following proof to show that for any choice of C and D and for each integer k 2 2,
b = bk - 1+ 6bk - 2'
Proof: Let C and D be any real numbers and let k be any integer such that k 2 2. By definition of bo, b,, b,, ...,
b -1 =
and
bk- 2 =
To complete the proof, you need to show that b, = b,, + 6b,2. When the expressions for b,- 1 and b, , are substituted into b,1 + 6b,
the result is
1
:- 2
| -•([
]) -[
bk - 1+ 6bk - 2 =
+ 6
]-«-a*-3).
= c| 3k-1
+D
Further simplification gives that
bk - 1+ 6bk - 2 =
which is what was to be shown.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa39a981d-f6af-4416-867c-0e8b7c81a8f4%2Fa769209c-d95f-4fa9-ac13-ce62325ab529%2F49quo5t_processed.png&w=3840&q=75)
Transcribed Image Text:Let bo, b,, b2, ... be the sequence defined by the explicit formula
b. = C. 3" + D(-2)" for every integer n 2 0,
where C and D are real numbers. Fill in the blanks in the following proof to show that for any choice of C and D and for each integer k 2 2,
b = bk - 1+ 6bk - 2'
Proof: Let C and D be any real numbers and let k be any integer such that k 2 2. By definition of bo, b,, b,, ...,
b -1 =
and
bk- 2 =
To complete the proof, you need to show that b, = b,, + 6b,2. When the expressions for b,- 1 and b, , are substituted into b,1 + 6b,
the result is
1
:- 2
| -•([
]) -[
bk - 1+ 6bk - 2 =
+ 6
]-«-a*-3).
= c| 3k-1
+D
Further simplification gives that
bk - 1+ 6bk - 2 =
which is what was to be shown.
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