$1,000 Invested at 6% for 3 Years
$1,196.68
Future Value (compounded monthly)
$1,000 invested at 6% annual compound interest (compounded monthly) for 3 years will grow to $1,196.68. You earn $196.68 in interest. At 6%, your money doubles in approximately 12 years (Rule of 72).
Year-by-Year Growth
| Year | Balance | Interest |
|---|---|---|
| 1 | $1,061.68 | $61.68 |
| 2 | $1,127.16 | $127.16 |
| 3 | $1,196.68 | $196.68 |
Quick Reference Table
| Principal | Rate | Years | Future Value |
|---|---|---|---|
| $1,000 | 4% | 3 yrs | $1,127.27 |
| $1,000 | 5% | 3 yrs | $1,161.47 |
| $1,000 | 7% | 3 yrs | $1,232.93 |
| $1,000 | 8% | 3 yrs | $1,270.24 |
| $1,000 | 6% | 1 yrs | $1,061.68 |
| $1,000 | 6% | 2 yrs | $1,127.16 |
| $1,000 | 6% | 5 yrs | $1,348.85 |
| $1,000 | 6% | 7 yrs | $1,520.37 |
| $1,000 | 6% | 10 yrs | $1,819.40 |
| $1,000 | 6% | 15 yrs | $2,454.09 |
Formula Used
A = P(1 + r/n)nt
- P = $1,000
- r = 6% = 0.06
- n = 12 (monthly)
- t = 3 years
- A = $1,196.68
Frequently Asked Questions
How much will $1,000 grow at 6% compound interest in 3 years?
$1,000 grows to $1,196.68. Interest earned: $196.68.
How long to double $1,000 at 6%?
Using the Rule of 72: 72 ÷ 6 ≈ 12 years.
What is the compound interest formula?
A = P(1 + r/n)^(nt). P=$1,000, r=6%=0.06, n=12, t=3.