1)- In an arithmetic series of 1, 2, 3, 4........... How many terms do you need to sum up to a 1,000,000?
2)-In the same series above, how many terms do you need to sum up to a 1,000,000, provided each subsequent term is increased by 1%, just as interest is paid on bank deposits?.
Thanks for any help.
+129849 1. We need to solve this :
[n (n + 1)] / 2 = 1,000,000
n^2 + n = 2,000,000
n^2 + n - 2,000,000 = 0
WolframAlpha gives the solution that n ≈ 1413.7......so......you need about 1414 terms
2. We need to solve this :
1 [ 1 - 1.01^n] [1 - 1.01] = 1000000
WolramAlpha gives the solution that n ≈ 925.6 ........so.....you need about 926 terms
1)-Using the formula: (F + (F + (N - 1) X D)) X 1/2 X N =S, where F=1st term, N=number of terms, D=difference between terms, S=Sum of all terms. We have:
(1+(1+(N-1) X 1)) X 1/2 X N=1,000,000, solving for N, we get:
Expand out terms of the left hand side:
N^2+N = 2000000
Add 1/4 to both sides:
N^2+N+1/4 = 8000001/4
Write the left hand side as a square:
(N+1/2)^2 = 8000001/4
Take the square root of both sides:
N+1/2 = (3 sqrt(888889))/2 or N+1/2 = -(3 sqrt(888889))/2
Subtract 1/2 from both sides:
N = (3 sqrt(888889))/2-1/2 or N+1/2 = -(3 sqrt(888889))/2
Subtract 1/2 from both sides:
Answer: |N = (3 sqrt(888889))/2-1/2 =~1,414 terms
2)- If each term is increased by 1%, then the cumulative terms become: 1, 1.01+1=2.01*1.01+1=3.0301, then this becomes:{[1+.01]^N-(.01N+1)} / .01^2=1,000,000
Solve for N:
N= 468.371=~469 terms
