For Maths Geniuses! can you simplify this?
t is a whole number and is greater than zero. It is a constant variable (this statement probably shows my ignorance in maths)
Any help would be appreciated
What is the solution??
or you can email sparky_3150@yahoo.com
Good luck, and thanks in advance! (This is not for school work BTW) ;)
Answer:
The answer is 2.5*1.4^t
You might also be interested in how to get there:
Begin with:
S = 1.4^(t-1) + 1.4^(t-2) + 1.4^(t-3)...+ 1.4^(t-n)
Factor out a 1.4^t:
S = (1.4^t)*(1.4^(-1) + 1.4^(-2) + 1.4^(-3) ... + 1.4^(-n))
Now define:
T(n) = 1.4^(-1) + 1.4^(-2) + 1.4^(-3) ... + 1.4^(-n)
Notice that
T(n)/1.4 = 1.4^(-2) + 1.4^(-3) + 1.4^(-4) ... + 1.4^(-(n + 1))
Also:
T(n) - 1.4(-1) + 1.4^(-(n + 1)) = 1.4^(-2) + 1.4^(-3) + 1.4^(-4) ... + 1.4^(-(n + 1))
These two expressions have the same right hand side so the left sides must be equal:
T(n)/1.4 = T(n) - 1.4(-1) + 1.4^(-(n + 1))
Solving for T(n):
T(n)/1.4 = T(n) - 1.4(-1) + 1.4^(-(n + 1))
T(n) = T(n)*1.4 - 1 + 1.4^(-n)
T(n) - T(n)*1.4 = -1 + 1.4^(-n)
0.4*T(n) = 1 - 1.4^(-n)
T(n) = 2.5*(1 - 1.4^(-n))
For large n, 1.4^(-n) goes to zero so:
T = 2.5 Substituting into S:
S = 1.4^t * T = 2.5*1.4^t
I agree with Pretzel's answer but this way might be simpler. Here's what I did:
1) Call the sum x.
2) x = 1.4^(t-1) + 1.4^(t-2) + 1.4^(t-3) + ...
3) x = (1/1.4)^(1-t) + (1/1.4)^(2-t) + (1/1.4)^(3-t) + ...
4) (1/1.4)^t * x = (1/1.4)^1 + (1/1.4)^2 + (1/1.4)^3 + ...
5) 1 + (1/1.4)^t * x = 1 + (1/1.4)^1 + (1/1.4)^2 + (1/1.4)^3 + ...
The right hand side of the equation is now an infinite geometric series of a number between 0 and 1, namely, 1/1.4. More specifically, this is the formula for the present value of a perpetuity (i.e., receive a dollar at the end of every year forever, and discount it back to the present).
Using the widely-known formula for a perpetuity, we proceed as follows:
5) 1 + (1/1.4)^t * x = 1/(1 - (1/1.4))
6)1.4^t + x = 1.4^t/(1-(1/1.4))
7) 1.4^t + x = 3.5 * 1.4^t
8) x = 2.5 * 1.4^t
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