In: Finance
Barbara is starting an investment program. She will invest $1,000 every year, with the first payment made today. She will make a total of 20 payments and will earn 6%. How much will she have at the end of 20 years?
FV =( P *( {(((1+R)^(N+1)) - 1) / R}-1) | |||||
=1000*(((((1+6%)^(20+1))-1)/6%)-1) | |||||
FV | 38,992.73 | ||||
Cross Check | |||||
Year | Period of Investment | Amount Invested | Amount with interest | ||
0 | 20 | 1,000 | 3,207.14 | =1000*(1+6%)^20 | |
1 | 19 | 1,000 | 3,025.60 | =1000*(1+6%)^19 | |
2 | 18 | 1,000 | 2,854.34 | =1000*(1+6%)^18 | |
3 | 17 | 1,000 | 2,692.77 | =1000*(1+6%)^17 | |
4 | 16 | 1,000 | 2,540.35 | =1000*(1+6%)^16 | |
5 | 15 | 1,000 | 2,396.56 | =1000*(1+6%)^15 | |
6 | 14 | 1,000 | 2,260.90 | =1000*(1+6%)^14 | |
7 | 13 | 1,000 | 2,132.93 | =1000*(1+6%)^13 | |
8 | 12 | 1,000 | 2,012.20 | =1000*(1+6%)^12 | |
9 | 11 | 1,000 | 1,898.30 | =1000*(1+6%)^11 | |
10 | 10 | 1,000 | 1,790.85 | =1000*(1+6%)^10 | |
11 | 9 | 1,000 | 1,689.48 | =1000*(1+6%)^9 | |
12 | 8 | 1,000 | 1,593.85 | =1000*(1+6%)^8 | |
13 | 7 | 1,000 | 1,503.63 | =1000*(1+6%)^7 | |
14 | 6 | 1,000 | 1,418.52 | =1000*(1+6%)^6 | |
15 | 5 | 1,000 | 1,338.23 | =1000*(1+6%)^5 | |
16 | 4 | 1,000 | 1,262.48 | =1000*(1+6%)^4 | |
17 | 3 | 1,000 | 1,191.02 | =1000*(1+6%)^3 | |
18 | 2 | 1,000 | 1,123.60 | =1000*(1+6%)^2 | |
19 | 1 | 1,000 | 1,060.00 | =1000*(1+6%)^1 | |
Total Amount | 38,992.73 |