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We know that the remainder Rn will satisfy |Rn| ⤠bn + 1 = 1 (n + 1)9n + 1. We must make n large enough so that this is less than 0.0001. Rounding to five decimal places, we have b2 = _________ , b3 =_________and b4 =__________

Sagot :

nuhulawal20

This question is incomplete, the complete question is;

We know that the remainder R[tex]_n[/tex] will satisfy | R[tex]_n[/tex] | ≤ b[tex]_{ n + 1[/tex] = 1 / ( n + 1 )9[tex]^{ n + 1[/tex].

We must make n large enough so that this is less than 0.0001.

Rounding to five decimal places,

we have b₂ = _________ , b₃ =_________and b₄ =__________

Answer:

b₂ = 0.00617, b = 0.00046 and  b₄ = 0.00004

Step-by-step explanation:

Given the data in the question;

| R[tex]_n[/tex] | ≤ b[tex]_{ n + 1[/tex] = 1 / ( n + 1 )9[tex]^{ n + 1[/tex]

Now,

b[tex]_{ n + 1[/tex] = 1 / ( n + 1 )9[tex]^{ n + 1[/tex]

b₂ = b[tex]_{ 1 + 1[/tex] = 1 / ( 1 + 1 )9[tex]^{ 1 + 1[/tex] = 1 / (2)9² = 1 / 162 = 0.00617   { 5 decimal places }

b₃ = b[tex]_{ 2 + 1[/tex] = 1 / ( 2 + 1 )9[tex]^{ 2 + 1[/tex] = 1 / (3)9³ = 1 / 2187 = 0.00046 { 5 decimal places }

b₄ = b[tex]_{ 3 + 1[/tex] = 1 / ( 3 + 1 )9[tex]^{ 3 + 1[/tex] = 1 / (4)9⁴ = 1 / 19683 = 0.00004 { 5 decimal places }

Therefore, b₂ = 0.00062, b = 0.00046 and  b₄ = 0.00004

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