Try to get them in a standard form, for which you probably saw a formula:
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.
can someone please help
i need to q5 and q6
i did attempt the questions, however i was getting incorrect answers.
any help is appreichated
thanks
regards
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
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]
= 3k /5 ------------------answer.
[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
S(5) = -73 ---------answer
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
= 6.72 --------------------answer.
Sorry, no more time, gotta go.
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)]
= 0.123 ----------------------answer.