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Tanzania [10]
3 years ago
7

Use the backward substitution method to obtain the closed form formula for the recurrence relation

Mathematics
1 answer:
abruzzese [7]3 years ago
3 0

Answer:

a_0=0; \; a_n=2^n-1 \; (n>0)

Step-by-step explanation:

a_0=0

a_1=2a_0+1=1

a_2=2a_1+1=3=1+2

a_3=2a_2+1=7=1+2+4

a_4=2a_3+1=15=1+2+4+8

a_5=2a_4+1=31=1+2+4+8+16

So, we can infer that

a_n=1+2^1+2^2+2^3+...+2^{n-1}

and we only need to find out a formula for this sum (a geometric reason with common ratio 2)

Let's call

S=1+2^1+2^2+...2^{n-1}

then,

2S=2+2^2+...2^{n-1}+2^n

hence,

S-2S=1-2^n

2S-S=S=2^n-1

and we have our closed formula for the sequence

a_0=0; \; a_n=2^n-1 \; (n>0)

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General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Order of Operations: BPEMDAS

<u>Algebra I</u>

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Step-by-step explanation:

<u>Step 1: Define</u>

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g(x) = x + 1

<u>Step 2: Find</u>

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  1. Substitute:                         (f - g)(x) = 4x - 2 - (x + 1)
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<em>Find (f - g)(3)</em>

  1. Substitute:                         (f - g)(3) = 3(3) - 3
  2. Multiply:                             (f - g)(3) = 9 - 3
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