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Write each series in sigma notation. The lower bound is given.


(1)/(2) + (1)/(4) +(1)/(8) + ...+(1)/(128) ;n=2

1 Answer

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Answer:

The sigma notation of the series is 8∑n=2 [1/2(1/2)^n-2]

Explanation:

* Lets revise the meaning of sigma notation

- A series can be represented in a compact form, called summation or

sigma notation.

- The Greek capital letter, ∑ , is used to represent the sum.

# Ex:

- The series 4 + 8 + 12 + 16 + 20 + 24 can be expressed 6Σ(n=1) 4n

- The expression is read as the sum of 4n as n goes from 1 to 6

- The coefficient of n is 4 because the common difference is 4

- The variable n is called the index of summation.

# Look to the attached figure to more understand

* Now lets solve the problem

∵ The series is 1/2 + 1/4 + 1/8 + ........... + 1/128 and n = 2

- There is a constant ratio between each to consecutive terms

∴ The series is geometric

∴ a1 = a , a2 = ar , a3 = ar² , a4 = ar³ , ..........

∵ an = a(r)^n-1, where a is the first term, r is the common ratio and n

is the position of the number in the series and the first n is 1

∵ 1/4 ÷ 1/2 = 12 , 1/8 ÷ 1/4 = 1/2

∴ The constant ratio is 1/2

∵ The first term is 1/2

∵ The last term is 1/128

∵ 1/128 = 1/2^7

∴ 1/128 = (1/2)^7

- There are seven terms in the sequence

∵ The n of the first term is 2

∴ The n of the last term = 7 + 1 = 8

* Now lets write the sigma notation

∴ 8∑n=2 [1/2(1/2)^n-2] ⇒ put them like the attached figure

- To generate the terms of the series given in sigma notation above,

replace n by 2 , 3 , 4 , 5 , 6 , 7 , 8

Write each series in sigma notation. The lower bound is given. (1)/(2) + (1)/(4) +(1)/(8) + ...+(1)/(128) ;n-example-1
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