geometric sums. 5 . derive the sum from n=0 to N for ar n =a((1-r N+1 )/(1-r)) b
ID: 2858166 • Letter: G
Question
geometric sums.
5. derive the sum from n=0 to N for arn=a((1-rN+1)/(1-r)) by
i. Notice that the sum from n=0 to N for arnis notation for a+ar+ar2+...+arN.. Denote Sn=a+ar+ar2+...+arn
ii. Write out rSn by explicityly multiplying both sides of the equation abve by r.
iii. Subtract both equations and solve for Sn
6. Note that we can write the future value off ALL of the payments as:
F = 100(1 + (r/12)240 + 100(1 + (r/12 ))239 + 100(1 + (r/12))238 + ... + 100(1 + (r/12)) 0
This is a geometric sum and we can use the formula in Problem 5 to evaluate it!
A. Write the above expression for F in summation notation. Then, use the formula from Problem 5 (and a bit of algebra) to show that:
F = (1200/r)[(1 + (r/12)241 1]
Explanation / Answer
5)
Terma are:
a,ar,ar^2,ar^3 and so on till ar^n
Sn=a+ar+ar^2+ar^3+.......+ar^n ....eqn 1
multiply both sides by r,
r*Sn=ar+ar^2+ar^23+ar^4+.......+ar^n+ ar^(n+1) ...eqn 2
subtract eqn1 from eqn 2
r*Sn-Sn = {ar+ar^2+ar^23+ar^4+.......+ar^n+ ar^(n+1)} - {a+ar+ar^2+ar^3+.......+ar^n }
Sn(r-1) = ar^(n+1) - a
Sn(r-1) = a(r^(n+1) -1)
Sn = a(r^(n+1) -1) / (r-1)
Sn = a(1 - r^(n+1)) / (1-r)
Proved
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