GEOMETRIC SERIES IN SIGMA NOTATION

Example 1 : 

Evaluate :

_{r = 1}∑^{r = 10} (2 ⋅ 4^r)

Solution :

Substitute the first few values into the formula. 

_{r = 1}∑^{r = 6} (2 ⋅ 4^r) = (2 ⋅ 4¹) + (2 ⋅ 4²) + (2 ⋅ 4³) + ......

= (2 ⋅ 4) + (2 ⋅ 16) + (2 ⋅ 64) + ......

= 8 + 32 + 128 + ......

The above is a geometric series with a₁ = 9 and d = 4.

Use S_n = a₁(1 - rⁿ)/(1 - r) to find the sum. 

= 8(1 - 4⁶)/(1 - 4)

= 8(1 - 4096)/(-3)

= 8(-4095)/(-3)

= 8(1365)

= 10920

Example 2 : 

Evaluate :

_{r = 3}∑^{r = 10} (3^{r - 1})

Solution :

Substitute the first few values into the formula. 

_{r = 3}∑^{r = 10} (3^{r - 1}) = 3^{3 - 1} + 3^{4 - 1} + 3^{5 - 1} + ......

= 3² + 3³ + 3⁴ + ......

= 9 + 27 + 81 + ......

The above is an arithmetic series with a₁ = 9 and d = 3.

Use S_n = a₁(1 - rⁿ)/(1 - r) to find the sum (from r = 3 to r = 10, there are 8 terms, so n = 8).

= 9(1 - 3⁸)/(1 - 3)

= 9(1 - 6561)/(-2)

= 9(-6560)/(-2)

= 9(3280)

= 29520

Example 3 : 

Write the following arithmetic series in sigma notation. 

4 + 8 + 16 + ........... + 1024

Solution :

In the given geometric series,

a₁ = 4

r = a₂/a₁ = 8/4 = 2

Find the formula for n^th term. 

a_n = a₁r^{n - 1}

Substitute a₁ = 4 and r = 2.

= 4(2)^{n - 1}

= 2²(2)^{n - 1}

= 2^{2 + n - 1}

= 2^{n + 1}

Let a_n = 1024.

a_n = 1024

2^{n + 1} = 1024

Write 1024 as a power of 2. 

2^{n + 1} = 2¹⁰

n + 1 = 10

n  =  9

n  =  9

In the given geometric series, 4 is the 1^st term and 1024 is the 9^th term. 

Then, 

4 + 8 + 16 + ........... + 1024  =  _{n = 1}∑^{n = 9} (2^{n + 1})

Example 4 : 

Write the following arithmetic series in sigma notation. 

9 + 3 + 1 + 1/3 ........... + 1/243

Solution :

In the given arithmetic series,

a₁ = 9

r = a₂/a₁ = 3/9 = 1/3

Find the formula for n^th term. 

a_n = a₁r^{n - 1}

Substitute a₁ = 9 and r = 1/3.

= 9(1/3)^{n - 1}

= 3²(3^-1)^{n - 1}

= 3²(3^{-n + 1})

= 3^{2 - n + 1}

= 3^{3 - n}

Let a_n = 1/243.

a_n = 1/243

3^{3 - n} = 1/243

Write 1/243 as a power of 3. 

3^{3 - n} = 1/(3⁵)

3^{3 - n} = 3^-5

3 - n = -5

-n = -8

n = 8

In the given geometric series, 9 is the 1^st term and 1/243 is the 8^th term. 

Then, 

9 + 3 + 1 + 1/3 ........... + 1/243 = _{n = 1}∑^{n = 8} (3^{3 - n})

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