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 =

This question is incomplete, the complete question is;

We know that the remainder R will satisfy | R | ≤ b = 1 / ( n + 1 )9.

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 =__________

b₂ = 0.00617, b = 0.00046 and  b₄ = 0.00004

Step-by-step explanation:

Given the data in the question;

| R | ≤ b = 1 / ( n + 1 )9

Now,

b = 1 / ( n + 1 )9

b₂ = b = 1 / ( 1 + 1 )9 = 1 / (2)9² = 1 / 162 = 0.00617   { 5 decimal places }

b₃ = b = 1 / ( 2 + 1 )9 = 1 / (3)9³ = 1 / 2187 = 0.00046 { 5 decimal places }

b₄ = b = 1 / ( 3 + 1 )9 = 1 / (4)9⁴ = 1 / 19683 = 0.00004 { 5 decimal places }

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

table 3 represents a direct variation function. : )

30,40,50 is correct answer