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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

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.2: Arithmetic Sequences
Problem 56E
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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.
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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