Question

Evaluate:
20
Σ 4(8/9)^n-1 = [?]
1
Round to the nearest hundredth.

Answer 1

General form of geometric progression

• ar^n-1

On comparing to the summation

• a=4
• r=8/9

Apply Sum formula

`\boxed{\sf S_n=(a(1-r^n))/(1-r)}`

`\\ \implies \sf S_(20)=(4\left(1-\left((8)/(9)\right)^(20)\right))/(1-\left((8)/(9)\right))`

`\\ \implies\sf S_(20)=32.586`

`\\ \sf\mplies S_(20)=32.59`

Answer 2

32.59 (nearest hundredth)

Explanation:

Geometric sequence

General form of a geometric sequence:
`a_n=ar^(n-1)`

(where a is the first term and r is the common ratio)

Given:

`\displaystyle \sum^(20)_(n=1) 4 \left((8)/(9)\right)^(n-1)`

Therefore:

• a = 4
• r = 8/9

Sum of the first n terms of a geometric series:

`S_n=(a(1-r^n))/(1-r)`

To find the sum of the first 20 terms, substitute the found values of a and r, together with n = 20, into the formula:

`\implies S_(20)=(4\left(1-\left((8)/(9)\right)^(20)\right))/(1-\left((8)/(9)\right))`

`\implies S_(20)=32.58609013...`

`\implies S_(20)=32.59\:\: \sf (nearest\:hundredth)`