How to do this?

N

nbkf1ab

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Nov 30, 2006
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7
Hello,

I have problem I cannot figure out:

a= 1855412.30469188*1.000711 = 1,856,732
b= 1,856,732*1.000711 = 1,858,052
c= 1,858,052*1.000711 = 1,859,373

a+b+c = 5,574,156


I need to find the formula to calcuate 5,574,156 in the final step.
** have tried (1855412.30469188*1.000711 ^3)*3 = 5578118.141 does not work


Thank you for all your help!
 
Although I have NO IDEA what you;re looking for, here is what I think you meant:

There's a common difference of 1320. For adding the first n terms, you have

\(\displaystyle \L \sum_{i=1}^{n} (1855412 + (i-1)(1320))\)

You can simplify this by using formulas for summation if you'd like (turns out to be a quadratic):

\(\displaystyle (n)(1855412) + (n)(n-1)(660) = 660n^2 + 1854752n\)

Your specific answer is when n=3.

-daon
 
THANKS

Hello,

Thanks for your help, but I am not getting the correct answer when I plug in the numbers:


(3*1855412.30469188)+3*(3-1)*660 = 5570196.914

It should equal this= 5,574,156


Any ideas?

Thanks!
 
nbkf1ab said:
It should equal this= 5,574,156
Click to expand...
Why "should" it equal this? What are you doing? What was the original question? How did you arrive at this stage?

Please be specific. Thank you.

Eliz.
 
Sorry for not giving enough detail:

The total is a+b+c from below. When trying to find complete this problem my knowns are:
- beginning amount: 1855412.30469188
- The the increase: 1.000711
- The increase needs to be cal from the peroid before.

The Unknown is the number of peroids you may have in the below I use 3 but they could go up to 52.


a= 1855412.30469188*1.000711 = 1,856,732
b= 1,856,732*1.000711 = 1,858,052
c= 1,858,052*1.000711 = 1,859,373

a+b+c = 5,574,156


thanks again for all the help
aaron
 
daon said:
Although I have NO IDEA what you;re looking for, here is what I think you meant:

There's a common difference of 1320. For adding the first n terms, you have

\(\displaystyle \L \sum_{i=1}^{n} (1855412 + (i-1)(1320))\)

You can simplify this by using formulas for summation if you'd like (turns out to be a quadratic):

\(\displaystyle (n)(1855412) + (n)(n-1)(660) = 660n^2 + 1854752n\)

Your specific answer is when n=3.

-daon
Click to expand...

My pattern seems correct with what you have posted, but I did forgot the extra factor (the 1.000711).

\(\displaystyle ((n)(1855412) + (n)(n-1)(660)) \cdot (1.000711)\)

That should be correct.

-daon
 
nbkf1ab said:
they could go up to 52.
Click to expand...
There could be a hint in there.

If one were to accumulate three weekly deposits of 1855412.30469188 with interest compounded weekly, using, say a little over 3.75% - perhaps 3.7650334%, one might get:

\(\displaystyle \L\;1855412.30469188*(1.000711^{1}+1.000711^{2}+1.000711^{3}) = 5574157\)

Generalizing to any value (n) up to 52, one might get:

\(\displaystyle \L\;1855412.30469188*\frac{1.000711\;-\;1.000711^{n+1}}{1\;-\;1.000711}\)
 
nbkf1ab said:
a= 1855412.30469188*1.000711 = 1,856,732
b= 1,856,732*1.000711 = 1,858,052
c= 1,858,052*1.000711 = 1,859,373
a+b+c = 5,574,156
Click to expand...

WHY are you starting with 1855412.30469188, then rounding to closest dollar?

If you're adding the results of the calculations, then this is what is being added:

a= 1,855,412.30469188*1.000711 = 1,856,731.50284051
b= 1,856,731.50284051*1.000711 = 1,858,051.63893903
c= 1,858,051.63893903*1.000711 = 1,859,372.71365432

a+b+c = 5,574,155.85543386

What you're TRYING to do is add the integer value of (calculation + .5).
NO formula will give you that!

Daon's 1320 common diff "appears" to work, but that's only because
the first 3 differences are: 1319.198..., 1320.136... and 1321.075...

Really, what you have is a savings account receiving .0711% weekly.
 
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