$1,000 Invested at 8% for 3 Years
$1,270.24
Future Value (compounded monthly)
$1,000 invested at 8% annual compound interest (compounded monthly) for 3 years will grow to $1,270.24. You earn $270.24 in interest. At 8%, your money doubles in approximately 9 years (Rule of 72).
Year-by-Year Growth
| Year | Balance | Interest |
|---|---|---|
| 1 | $1,083.00 | $83.00 |
| 2 | $1,172.89 | $172.89 |
| 3 | $1,270.24 | $270.24 |
Quick Reference Table
| Principal | Rate | Years | Future Value |
|---|---|---|---|
| $1,000 | 6% | 3 yrs | $1,196.68 |
| $1,000 | 7% | 3 yrs | $1,232.93 |
| $1,000 | 9% | 3 yrs | $1,308.65 |
| $1,000 | 10% | 3 yrs | $1,348.18 |
| $1,000 | 8% | 1 yrs | $1,083.00 |
| $1,000 | 8% | 2 yrs | $1,172.89 |
| $1,000 | 8% | 5 yrs | $1,489.85 |
| $1,000 | 8% | 7 yrs | $1,747.42 |
| $1,000 | 8% | 10 yrs | $2,219.64 |
| $1,000 | 8% | 15 yrs | $3,306.92 |
Formula Used
A = P(1 + r/n)nt
- P = $1,000
- r = 8% = 0.08
- n = 12 (monthly)
- t = 3 years
- A = $1,270.24
Frequently Asked Questions
How much will $1,000 grow at 8% compound interest in 3 years?
$1,000 grows to $1,270.24. Interest earned: $270.24.
How long to double $1,000 at 8%?
Using the Rule of 72: 72 ÷ 8 ≈ 9 years.
What is the compound interest formula?
A = P(1 + r/n)^(nt). P=$1,000, r=8%=0.08, n=12, t=3.