1 more limiting sum question (1 Viewer)

[atakach99]

Evaluate

the sum (sigma notation) going to infinity starting from 3 of 5(1/3)^(n-1)

whenever i do it i get the answer as 5/6 but in the back of the book the answer is 1 1/4

plz help

btw this is Ex.8.3 Q)12)c)
Book name: MAths in focus bk 2 extention 1

 

[lyounamu]

Evaluate

the sum (sigma notation) going to infinity starting from 3 of 5(1/3)^n-1

whenever i do it i get the answer as 5/6 but in the back of the book the answer is 1 1/4

plz help

btw this is Ex.8.3 Q)12)c)
Book name: MAths in focus bk 2 extention 1

Strange.. I got the same answer as you... It seem right though. If I get it right, I will repost immediately

EDIT: 5/6 seems to be the right answer.

 

[Slidey]

Going to guess that's ^(n-1).

Lim to infinity of Sum 5*(1/3)^(n-1) starting at 3 =
5/9 * 1/(2/3) = 5/6

For some reason they gave an answer that starts at n=2 instead of 3, and uses the completely incorrect formula a/(1+r).

 

[atakach99]

ohh ok so the answer in the back of the book is probably wrong..if anyone has the book can u plz check the question urself

 

[conics2008]

hey buddy, the answer is right, you're looking at a different solution. thats the solution for 12 a lol... xD

 

[Perspesutastrum]

Ok, so I just need to clarify... a limiting sum is just the sum of all terms in a geometric sequence, and to infinity.
For example, if I have a sequence in which a=2 and r=0.5, then the limiting sum would be
S = 2 / 1-0.5
S approaches 4

...?

And if someone asks if a limiting sum exits for a particular sequence, and -1<r>1 , then the limiting sum does exist?

Why is it called a limiting sum anyway?

(I apologise if I'm using the wrong words, but this part of the sequences topic really confused me.)

 

[lyounamu]

Ok, so I just need to clarify... a limiting sum is just the sum of all terms in a geometric sequence, and to infinity.
For example, if I have a sequence in which a=2 and r=0.5, then the limiting sum would be
S = 2 / 1-0.5
S approaches 4

...?

And if someone asks if a limiting sum exits for a particular sequence, and -1<R>1 , then the limiting sum does exist?

Why is it called a limiting sum anyway?

(I apologise if I'm using the wrong words, but this part of the sequences topic really confused me.)

A limiting sum is just the sum of all term when a geometric sequences approaches 0 (therefore -1 < r < 1).

To see if a limiting sum exists, just find the r, if the absolute value of r is larger than 1 (i.e. r<-1 or r>1), there is no limiting sum.

Limiting sum only exists when the absolute value of r is less than 1.

 

[tacogym27101990]

hey buddy, the answer is right, you're looking at a different solution. thats the solution for 12 a lol... xD

12c) 's answer is also 1.25 in my book.
so it is wrong yeah??