SOLUTION: 1.1 Given 16(x-2)^3+8(x-2)^4+4(x-2)^5+...... As a geometric sequence and x is not equal to 2 1.1.1 Calculate the value of x for which sequence converges 1.1.2 Determine the s

Algebra ->  Sequences-and-series -> SOLUTION: 1.1 Given 16(x-2)^3+8(x-2)^4+4(x-2)^5+...... As a geometric sequence and x is not equal to 2 1.1.1 Calculate the value of x for which sequence converges 1.1.2 Determine the s      Log On


   

Question 626875: 1.1 Given 16(x-2)^3+8(x-2)^4+4(x-2)^5+...... As a geometric sequence and x is not equal to 2
1.1.1 Calculate the value of x for which sequence converges
1.1.2 Determine the sum to infinity of the series if x=2.5
Answer by reviewermath(1029) About Me  (Show Source):
You can put this solution on YOUR website!
1.1 To get the next term, multiply by %281%2F2%29%2A%28x-2%29. This is the common ratio, r. The first term is a%5B1%5D+=+16%28x-2%29%5E3. The sum of the infinite geometric series is equal to
S+=+a%5B1%5D%2F%281-r%29

16%28x-2%29%5E3%2B8%28x-2%29%5E4%2B4%28x-2%29%5E5%2B... = 16%28x-2%29%5E3%2F%281-%281%2F2%29%2A%28x-2%29%29+
= 32%28x-2%29%5E3%2F%284-x%29%29+
1.1.1 The geometric series is convergent iff abs%28r%29+%3C+1
abs%28%281%2F2%29%2A%28x-2%29%29%3C1
+-1+%3C+%281%2F2%29%2A%28x-2%29+%3C+1
-2+%3C+x-2+%3C+2
highlight%280%3Cx%3C4%29
1.1.2 x=2.5
32%28x-2%29%5E3%2F%284-x%29%29+, substitute x = 2.5
= 32%282.5-2%29%5E3%2F%284-2.5%29%29+
=highlight%288%2F3%29