Geometric Series Q (1 Viewer)

[sae24]

Hi, For this question, for part b,
Would you substitute a=2(0.9) or a=1 ; am very confused as different answer sources substituted either one.

Please help, thank you !

 

[aqwerty13402]

you can do both. You can leave one on the outside, and form the GP with everything after that. So it'd be

1 + 2(0.9 + 0.9^2 + 0.9^3...) a =0.9 r= 0.9 n= n
Sn = 1 + 2((0.9(0.9^2-1)/0.9)

The key if u leave the 1 out the front and start the GP after the one, is to factorise out the 2. That wasy you have a clear GP inside

 

[cossine]

View attachment 43303

Hi, For this question, for part b,
Would you substitute a=2(0.9) or a=1 ; am very confused as different answer sources substituted either one.

Please help, thank you !

a is always the first term in the series. So therefore a = 1. r is defined to be the ratio of successive terms. So r = 2(0.9)/1 = 2(0.9)

 

[cossine]

you can do both. You can leave one on the outside, and form the GP with everything after that. So it'd be

1 + 2(0.9 + 0.9^2 + 0.9^3...) a =0.9 r= 0.9 n= n
Sn = 1 + 2((0.9(0.9^2-1)/0.9)

The key if u leave the 1 out the front and start the GP after the one, is to factorise out the 2. That wasy you have a clear GP inside

This is not correct. You have changed series.

Also the formula for geometric series has not been applied correctly. You should power to n. Also number of terms are now n-1 since you have remove the 1.

 

[aqwerty13402]

This is not correct.

 

[liamkk112]

a is always the first term in the series. So therefore a = 1. r is defined to be the ratio of successive terms. So r = 2(0.9)/1 = 2(0.9)

but if r = 2(0.9), then the powers of 2 would have to increase across the series because each time u would also have to multiply by 2

i think the best method is this:
write Sn = 1 + 2 (0.9) + 2 (0.9)^2 + ... + 2(0.9)^n

now we can take a= 2(0.9) to be the first term, and the ratio to be r = 0.9; the 1 is just an additional term we need to add on. now there are n-1 terms, so by the GP sum formula,



this makes perfect sense since when n = 1, S_n = 1, and when n> 1 the result still matches Sn = 1 + 2 (0.9) + 2 (0.9)^2 + ... + 2(0.9)^n

 

[liamkk112]

but if r = 2(0.9), then the powers of 2 would have to increase across the series because each time u would also have to multiply by 2

i think the best method is this:
write Sn = 1 + 2 (0.9) + 2 (0.9)^2 + ... + 2(0.9)^n

now we can take a= 2(0.9) to be the first term, and the ratio to be r = 0.9; the 1 is just an additional term we need to add on. now there are n-1 terms, so by the GP sum formula,



this makes perfect sense since when n = 1, S_n = 1, and when n> 1 the result still matches Sn = 1 + 2 (0.9) + 2 (0.9)^2 + ... + 2(0.9)^n

something to note is what n means in S_n; if n is defined as the time to reach the ground from the height above the ground, then the number of terms is n-1, which again would make sense since S_1 = 1. but if n is defined as the time to complete one "bounce", meaning up and then back down after hitting the ground, then the number of terms would be n, because we wouldn't count the initial drop as a bounce. it's probably a good idea to indicate the reasoning for which number of terms you pick, but in the limit as n goes to infinity they obviously produce the same result

 

[cossine]

but if r = 2(0.9), then the powers of 2 would have to increase across the series because each time u would also have to multiply by 2

i think the best method is this:
write Sn = 1 + 2 (0.9) + 2 (0.9)^2 + ... + 2(0.9)^n

now we can take a= 2(0.9) to be the first term, and the ratio to be r = 0.9; the 1 is just an additional term we need to add on. now there are n-1 terms, so by the GP sum formula,



this makes perfect sense since when n = 1, S_n = 1, and when n> 1 the result still matches Sn = 1 + 2 (0.9) + 2 (0.9)^2 + ... + 2(0.9)^n

Ah I missed that detail