Amount with compounding
Multiply by the same number every year. A power is a short way to write it.
A quick way to the amount
₹6000 is in an FD at 10% p.a. The interest is added back every year, for 3 years.
Which calculation gives the amount after 3 years?
What this lesson covers
The idea
With compounding, the amount after t terms is p × (1 + r) × (1 + r) × … × (1 + r) = p(1 + r)^t, where p is the principal and r the rate of interest per term.
A quick way to the amount
₹6000 is in an FD at 10% p.a. The interest is added back every year, for 3 years.
Which calculation gives the amount after 3 years?
- 6000 × 1.1 × 1.1 × 1.1
- 6000 × 1.3
- 6000 × 1.1 × 3
Multiply, year after year
Each bar is the amount at the end of a year. Every year the amount is multiplied by the same number. Choose the years and the rate, and reach the amounts in the goal.
The same multiplier every year
With compounding, the amount after t terms is p × (1 + r) × (1 + r) × … × (1 + r) = p(1 + r)^t, where p is the principal and r the rate of interest per term.
Each year the amount grows by 10% of itself, so it becomes 1.1 times as much: After 1 year: 6000 × 1.1 After 2 years: 6000 × 1.1 × 1.1 After 3 years: 6000 × 1.1 × 1.1 × 1.1 = 6000 × 1.331 = ₹7986
Multiplying 1.1 by itself t times is written 1.1^t, so after t years the amount is 6000 × 1.1^t. In general it is p × (1 + r)^t.
Compare: without compounding the 3 years add the interest three times, 1 + 0.1 + 0.1 + 0.1 = 1.3. With compounding they multiply: 1.1 × 1.1 × 1.1 = 1.331.
Notes
With compounding, the amount after t terms is p × (1 + r) × (1 + r) × … × (1 + r) = p(1 + r)^t, where p is the principal and r the rate of interest per term.
Check yourself
₹2000 is put in an FD at 10% p.a. with the interest added back every year. What is the amount after 2 years (in ₹)?
Answer: 2420
2000 × 1.1 = 2200 2200 × 1.1 = ₹2420
₹5000 is put in an FD at 20% p.a. with the interest added back every year. What is the amount after 2 years (in ₹)?
Answer: 7200
5000 × 1.2 = 6000 6000 × 1.2 = ₹7200
Which expression gives the amount after t years with compounding, for a principal p at a rate r per year?
By what number is ₹6000 multiplied to get the amount after 3 years at 10% p.a., compounded? Give the number.
Answer: 1.331
1.1 × 1.1 = 1.21 1.21 × 1.1 = 1.331
- p × (1 + r)^t — correct. Yes! The amount is multiplied by (1 + r) every year, t times.
- p × (1 + r × t). That adds r once for each year (1 + r + r + …). It is the amount without compounding.
- p × (1 + r) × t. That multiplies by (1 + r) only once and then by t. The multiplier (1 + r) must be used t times: (1 + r)^t.