+579 I need to know if there was a better way to do this involving sigma.
To save money for a vacation, you set aside 100 dollars. for each month thereafter, you plan to set aside 10% more than the last month for 12 months. how much money will you have saved up after the 12 months?
i painstakenly calculated all 12 numbers.
100 + 110 + 121 + 133.1 +146.41 + 161.051 + 177.1561 + 194.87171 + 214.358881 + 235.7947691 + 259.37424601 + 285.311670611 = 285.311670611
was there a better and more time saving way to do this?
i would like an answer please :(
Well, there is a financial formula for that: FV=P{[1 + R]^N - 1/ R}, Where R=Interest rate per period, N=number of periods, P=periodic payment, FV=Future value.
FV = $100 x {[1 + 10%]^12 - 1 / R}
FV = $100 x {[1 + 0.10]^12 -1 / 0.10}
FV = $100 x {[1.10]^12 - 1 / 0.10}
FV = $100 x {[3.138428376721 - 1 / 0.10}
FV = $100 x {2.138428376721 -1/ 0.10}
FV = $100 x 21.38428376........
FV = $2,138.43
+579 Yes i know the answer, i was wondering if there was a sigma equation to show this
+118608 ∑[100*1.1^n, n, 0, 11] =$2,138.43
Are you saying you can do it like this on a calculator guest?
What calculator are you referring to?
Yes Melody. On my personal calculator you can, because it has the "Sigma" notation built into it. That is exactly how I got that answer.
+36916 I think this states you start with 100 then for 11 more months add 10% total 12 months correct?
The question states: Start with 100 then for 12 more months adds 10 %
so would it be n,0,12 ?
Yea, it sums them up like this:
$100*1.1^0 + $100*1.1^1 + $100*1.1^2 + $100*1.1^3..............$100*1.1^11 =$2,138.43.
Because the first exponent MUST be zero, and the 12th payment would be 11..............., for this:
$100*1.1^0 =$100 x 1 =$100, which is the first payment......and so on.
+36916 OK...I think the question is worded incorrectly..... it STARTS with 100 then for TWELVE MORE months 10% is added......but I think it MEANT 11 MORE months (for a total of 12 months)
EP: The reason for the notation{n, 0, 11} is because in financial investments, the first payment is taken to be made at the END of the period, in this case the END of the first month. Example: First pmt. made on Jan.31 of $100. Now, you will wait for the whole of Feb. to get interest on it. So, the END of Feb. you will have: $100 x 1.1 =$110. Then you add another $100 deposit =$210 at the END of Feb. This has exactly the same outcome as the young man calculated in his question: $100, $110, $121....etc.
+36916 Thanx, I understand that.....BUT, I was merely pointing out that the question is worded improperly.
you start out with 100 and then for 12 MORE months you add 10% (per the question)
'you set aside 100 dollars
'for each month thereafter, you plan to set aside 10% more than the last month for 12 months
So it is a series of 13 payments essentially....first one 100 then twelve more.
It is like saying , in February , I put 100 in the bank....then starting in July I made a deposit of 110
(1st increasing deposit) ...then I did this ELEVEN MORE times for a total of TWELVE more deposits after the 100..... Do you see what I am trying to say? ' for each month thereafter ....for twelve months'
I see in the poster's answer he only made increasing deposits ELEVEN more times....so what the question should say is you start with 100 then for ELEVEN months you add 10% to the previous month's deposit.
Anyway....minor point. Just how I read the question.....but apparently no one else did, so I'll go with the flow......
Thanx .... G'Day !
+118608 100 + 110 + 121 + 133.1 +146.41 + 161.051 + 177.1561 + 194.87171 + 214.358881 + 235.7947691 + 259.37424601 + 285.311670611 = 285.311670611
these are the terms of a GP
r= 1.1
this is the best way to present it using sigma notation but it does not give you the answer.
\(\displaystyle \sum_{n=0}^{11} 100*101^n\)
If you want the answer it is easier to do it as the sum of a GP (that is if you have learned GPs yet)
a=100
n=12 (because n=1 is always the first one)
r=1.1
\(S_{n}=\frac{a(r^n-1)}{r-1}\\ S_{12}=\frac{100(1.1^{12}-1)}{1.1-1}\\ S_{12}=\frac{100(1.1^{12}-1)}{0.1}\\ S_{12}=1000(1.1^{12}-1)\\ \)
1000(1.1^12-1) = 2138.428376721
If I take your addition and put it into the web2 calc I get exactly the same answer.
2138.428376721 = 2.138428376721e3 = 2138.428
+579 wow i leave a question overnight and get answers galore. thanks guest for giving me the formula
