Question

Can some one help me with geometric sequence?​

Answer 1

See below

Explanation:

We have the series

`$\sum _( n=0 )^( \infty ){ 3(0.8)^n}$`

(a) To show that the sum of first
`n` terms is
`S_n= 15(1-0.8^n)`

Consider that the this sum is given as

`S_n = 3\frac { 1-{ r }^n}{ 1-r }`

In this case, the ratio is 0.8, thus,
`r=0.8`

`S_n = 3\frac { 1-{ 0.8 }^n }{ 1-0.8 }`

`3\frac { 1-{ 0.8 }^( n ) }{ 1-0.8 } = 15(1-0.8^n) \iff 3(1-{ 0.8 }^( n )) = 15(1-0.8^n)( 1-0.8) \iff 3 = 15(1-0.8) \iff 3 = 15-12 = 3`

Therefore,

`S_n = 3\frac { 1-{ 0.8 }^n }{ 1-0.8 } = 15(1-0.8^n)`

`(S_6)/(S_4) = (15(1-0.8^6))/(15(1-0.8^4)) = ((1-0.8^6))/((1-0.8^4)) = (0.737856)/(0.5904) = (737856)/(5904) = (144\cdot 5124)/(144\cdot 41) = \boxed{(5124)/(41) }`

(b) Calculate the smallest integer value of
`n` such that
`S_(40) -S_n < (1)/(2)`

Once
`S_(40) \approx 15 \text{ and } S_n>14.5, \text{ thus } S_(40) \approx S_n, \text{ but } n \text{ is considerably smaller than 40}`

I am considering
`S_(40)=15`, so

`15-15(1-0.8^n) < (1)/(2) \implies 0.8^n < 0.03 \implies n> 15.71`

`\text{Once } n\in\mathbb{Z}, \text{ then } n=16`