sequences and series help

quickquestion805

15

posting question and mark scheme below this. question b.

why is the power n rather than n-1?

OP

Morning.

An example:
After 1 year, it would increase by 2%. So would be 1.02(^1).
If it was n-1 and not n then you'd be doing 1.02(^0) and anything to the power of zero is 1, so it wouldn't make sense.

Reply 3

quickquestion805

OP

15

Original post by mesub

Morning.

An example:
After 1 year, it would increase by 2%. So would be 1.02(^1).
If it was n-1 and not n then you'd be doing 1.02(^0) and anything to the power of zero is 1, so it wouldn't make sense.

I'm still a little confused. Isn't the first term in the sequence 25000 x 1.02^0? or was it 25000 x 1.02 for the year 2013? if that were the case wouldn't a=25000 x 1.02? sorry if I'm being an idiot right now.

Reply 4

mqb2766

21

Original post by quickquestion805

I'm still a little confused. Isn't the first term in the sequence 25000 x 1.02^0? or was it 25000 x 1.02 for the year 2013? if that were the case wouldn't a=25000 x 1.02? sorry if I'm being an idiot right now.

After n=0 years (beginning 2012) the population is 25000, after n=1 years (beginning 2013) the popullation is 25000*1.02, after n=2 .... so
p = 25000*1.02^n
where n is the number of years.

Note you could have modelled it as
25000*1.02^(n-1)
with n=1 being the beginning of 2012 .... but when you solved for n, youd have to subtract 1 to it to get the number of years or equivalently solve for "n-1". However, its easier to set the problem up with the exponent being n and solve for n so you dont have to fiddle around with offsets.

Its obviously equivalent to a compound interest problem, so year 0 is your initial balance, after n=1 years you have 1 years interest, after n=2 years you have 2 years interest etc.

(edited 9 months ago)

sequences and series help

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