Question

Can some one help me with geometric sequence?​

Answer 1

Answer:

See below

Step-by-step 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

$S_n = 3\dfrac { 1-{ r }^n}{ 1-r }$

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

$S_n = 3\dfrac { 1-{ 0.8 }^n }{ 1-0.8 }$

$3\dfrac { 1-{ 0.8 }^{ n } }{ 1-0.8 } = 15(1-0.8^n) \iff 3(1-{ 0.8 }^{ n }) = 15(1-0.8^n)( 1-0.8) \iff 3 = 15(1-0.8) \iff 3 = 15-12 = 3$

Therefore,

$S_n = 3\dfrac { 1-{ 0.8 }^n }{ 1-0.8 } = 15(1-0.8^n)$

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

(b) Calculate the smallest integer value of $n$ such that $S_{40} -S_n < \dfrac{1}{2}$

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

$15-15(1-0.8^n) < \dfrac{1}{2} \implies 0.8^n < 0.03 \implies n> 15.71$

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