Find the sum 0.000061 + 0.000061(1.25) + 0.000061(1.25)^2 + ... + 0.000061(1.25)^9.

Step-by-step solution

1. 1

   Recognise the structure. Every term is the previous one multiplied by $1.25$, so this is a geometric series. Reading off its parameters: first term $a=0.000061$, common ratio $r=1.25$, and — counting exponents $0$ through $9$ — exactly $n=10$ terms. Miscounting $n$ as $9$ is the usual trap.

2. 2

   Factor out the common first term.

   $0.000061\left(1+1.25+1.25^{2}+\cdots+1.25^{9}\right)$

   This leaves a clean series with first term $1$, so only the bracket needs the formula.

3. 3

   Apply the finite geometric sum formula. For $r\neq1$,

   $S_{n}=\frac{r^{n}-1}{r-1}=\frac{1.25^{10}-1}{1.25-1}=\frac{1.25^{10}-1}{0.25}=4\left(1.25^{10}-1\right)$

   Dividing by $0.25$ is the same as multiplying by $4$ — a useful simplification that keeps the arithmetic exact.

4. 4

   Evaluate the power. Since $1.25=\tfrac54$,

   $1.25^{10}=\frac{5^{10}}{4^{10}}=\frac{9\,765\,625}{1\,048\,576}=9.313225746\ldots$

   so the bracket equals $4(9.313225746-1)=4(8.313225746)=33.252902985$.

5. 5

   Multiply by the first term — carefully.

   $0.000061\times33.252902985=0.0020284271$

   Working the whole calculation in exact fractions gives $\dfrac{531\,739\,989}{262\,144\,000\,000}=0.00202842708\ldots$ ✓, confirming the decimal. Note the answer is $0.0020284$, not $0.0020289$ — a slip in the fourth significant figure is easy to make here and changes the result by about $0.025\%$.

6. 6

   Sanity-check the magnitude. Ten terms averaging roughly $0.0002$ each should total around $0.002$ ✓. The growth factor also checks out: the last term is $0.000061\times1.25^{9}\approx0.000455$, about $7.5$ times the first — consistent with a ratio above $1$ compounding nine times.