1. ## geometric series help.

hello

i am currently studying geometric series. and i have been trying to do some questions which i just can't seem to solve. i have scanned the questions and have attached them to this thread.

i need to q5 and q6

i did attempt the questions, however i was getting incorrect answers.

any help is appreichated

thanks

regards

2. Try to get them in a standard form, for which you probably saw a formula:

$\displaystyle \sum\limits_{r = 1}^\infty {\frac{{3k}}{{6^r }}} = 3k\sum\limits_{r = 1}^\infty {\left( {\frac{1}{6}} \right)^r } = 3k\left( {\sum\limits_{r = 0}^\infty {\left( {\frac{1}{6}} \right)^r - 1} } \right) = 3k\left( {\frac{1}{{1 - \frac{1}{6}}} - 1} \right) = \frac{3}{5}k$

3. when u mean formula do mean

a(r^n-1)
----------
r-1

and

a(1-r^n)
----------
1-r

?

thanks

regards

4. For an infinite sum, we even have something simpler. I used:

$\displaystyle \sum\limits_{r = 1}^\infty {a^r } = \frac{1}{{1 - a}}$

5. oh ok.

can you please just help me out on any question from q6 please

thanks

regards

6. Notice that:

$\displaystyle \frac{9}{4} = \left( {\frac{2}{3}} \right)^{ - 2} ,\frac{3}{2} = \left( {\frac{2}{3}} \right)^{ - 1} ,0 = \left( {\frac{2}{3}} \right)^0 , \cdots ,\frac{{1024}}{{59019}} = \left( {\frac{2}{3}} \right)^{10}$

So if we shift the exponent by 2, we start at exponent 0 and go to 12.
We can then write this finite geometrical sum compactly, as follows:

$\displaystyle S = \sum\limits_{n = - 2}^{10} {\left( {\frac{2}{3}} \right)^n } = \frac{{\left( {\frac{2}{3}} \right)^{ - 2} - \left( {\frac{2}{3}} \right)^{11} }}{{1 - \frac{2}{3}}} = \frac{{1586131}}{{236196}} \approx 6.72$

7. when i was doing this i was using logs.

but i ended up with wrong answer

TD. do you think i should do exponentials and logarithms before attempting geometric sequences ?

8. Do you understand what I did? And do you recognise the formula I used?

9. the formula you use was

a(1-r^n)
----------
1-r

am i right in saying this ?

thanks

regards

10. Almost, that's for when your index starts at 0, now we began at -2.
The first 1 in the numerator is now r^q instead of 1, where q is this start index.

11. is there a way of doing q6 with logarithms

the reason why i ask is because the examples i came accross used logs.
and the formula used the answer above is one that i am unaware of and also its not in my textbook

however the answer is correct though.

many thanks

regards

12. Originally Posted by rpatel
hello

i am currently studying geometric series. and i have been trying to do some questions which i just can't seem to solve. i have scanned the questions and have attached them to this thread.

i need to q5 and q6

i did attempt the questions, however i was getting incorrect answers.

any help is appreichated

thanks

regards
Here is one way.

[Q5,a] Summation of [3k / 6^n] from n=1 to n=infinity.
I changed r into n because I want to maintain r = ratio
In series form, that is
3k/(6^10 +3k/(6^2) +3k/(6^3) +....+3k/(6^infinity)

r = a2/a1 = 6^(-1) = 1/6

It is a geometric series where r = 1/6
So,
Sn = (a1)[(1-r^n)/(1-r)]
n = infinity
S(infinity) = (3k /6)[(1 -(1/6)^infinity) / (1 -1/6)]
= (3k /6)[(1-0) / (5/6)]
= (3k /6)[6/5]

[Q5,b] Summation of [(-2)^n +n -2] from n=3 to n=7.
Split that into a geometric and an arithmetic series.
Sn = (a1)[(1-r^n)/(1-r)] +(n/2)[b1 +bn]

In the geometric series (-2)^n, the r is (-2).
In the arithmetic series (n-2), the d is 1.

n=3 to n=7 is 5 elements.

So,
S(5) = (-2)^3 *[(1 -(-2)^5)/(1-(-2))] +(5/2)[(3-2) +(7-2)]
S(5) = (-8)[(1-(-32)/(1+2)] +(2.5)[6]
S(5) = -88 +15

Note : S(infinity) and S(5) read "S sub infinity" and "S sub 5".

-----------------------------------------------
[Q6,a]

9/4 +3/2 +1 +.....+1024/59,049

A geometric series.
r = (3/2)/(9/4) = 2/3

a1 = 9/4 = (3/2)^2 = (2/3)^(-2)
an = 1024/59,049 = (2/3)^10
So from -2 to 10. That's 13 elements.

S(13) = (9/4)[(1 -(2/3)^13) / (1 -(2/3)]
= (9/4)[(0.99486)/(0.33333)]
= 6.71537

Sorry, no more time, gotta go.

13. Originally Posted by rpatel
hello

i am currently studying geometric series. and i have been trying to do some questions which i just can't seem to solve. i have scanned the questions and have attached them to this thread.

i need to q5 and q6

i did attempt the questions, however i was getting incorrect answers.

any help is appreichated

thanks

regards
Let me finish this question.

[Q6,b]
1/5 -3/20 +9/80 -....+6561/327,680

r = (-3/20) / (1/5) = (-3/20)*(5/1) = -3/4
r = (9/80) /(-3/20) = (9/80)*(-20/3) = -3/4

Okay, so the series is geometric, with r = -3/4 and a1 = 1/5

Rewriting the series,
(1/5)(-3/4)^0 +(1/5)(-3/4)^1 +(1/5)(-3/4)^2 +....+(1/5)(-3/4)^8
From 0 to 8, so there are 9 elements, or n=9.

S(9) = (1/5)[(1 -(-3/4)^9) / (1 -(-3/4))]
= (1/5)[(1.075)/(1.75)]