Question

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

Answer 1

Answer:

32.59 (nearest hundredth)

Step-by-step 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:

$\displaystyle \sum^{20}_{n=1} 4 \left(\dfrac{8}{9}\right)^{n-1}$

Therefore:

• a = 4
• r = 8/9

Sum of the first n terms of a geometric series:

$S_n=\dfrac{a(1-r^n)}{1-r}$

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

$\implies S_{20}=\dfrac{4\left(1-\left(\frac{8}{9}\right)^{20}\right)}{1-\left(\frac{8}{9}\right)}$

$\implies S_{20}=32.58609013...$

$\implies S_{20}=32.59\:\: \sf (nearest\:hundredth)$                    

Answer 2

General form of geometric progression

• ar^n-1

On comparing to the summation

• a=4
• r=8/9

Apply Sum formula

$\boxed{\sf S_n=\dfrac{a(1-r^n)}{1-r}}$

$\\ \implies \sf S_{20}=\dfrac{4\left(1-\left(\frac{8}{9}\right)^{20}\right)}{1-\left(\frac{8}{9}\right)}$

$\\ \implies\sf S_{20}=32.586$

$\\ \sf\mplies S_{20}=32.59$