Question

If \(X=\dfrac{a}{1+r}+\dfrac{a}{(1+r)^2}+\cdots+\dfrac{a}{(1+r)^n}\), then what is the value of \(a+a(1+r)+a(1+r)^2+\cdots+a(1+r)^{n-1}\)?

• A. \(X\left[(1+r)+(1+r)^2+\cdots+(1+r)^n\right]\)
• B. \(X(1+r)^n\)
• C. \(X\cdot\dfrac{(1+r)^n-1}{r}\)
• D. \(X(1+r)^{n-1}\)

Hint:

Multiply X by \((1+r)^n\) and notice each term's power of \((1+r)\) shifts to match the terms of the target series exactly.

Answer 1

The Correct Option is B

Step 1: Recognize X as a geometric series and sum it.
\[ X=\frac{a}{1+r}+\frac{a}{(1+r)^2}+\cdots+\frac{a}{(1+r)^n}=a\sum_{k=1}^{n}\left(\frac{1}{1+r}\right)^k \]
This is a geometric series with first term \(\frac{a}{1+r}\), common ratio \(\frac{1}{1+r}\), and \(n\) terms. Using the standard sum formula:
\[ X=\frac{a}{1+r}\cdot\frac{1-\left(\frac{1}{1+r}\right)^n}{1-\frac{1}{1+r}} \]

Step 2: Simplify X.
Since \(1-\dfrac{1}{1+r}=\dfrac{r}{1+r}\), we get
\[ X=\frac{a}{1+r}\cdot\left[1-(1+r)^{-n}\right]\cdot\frac{1+r}{r}=\frac{a}{r}\left[1-(1+r)^{-n}\right] \]

Step 3: Sum the target series, Y.
Let \(Y=a+a(1+r)+a(1+r)^2+\cdots+a(1+r)^{n-1}\), a geometric series with first term \(a\), common ratio \((1+r)\), and \(n\) terms:
\[ Y=a\cdot\frac{(1+r)^n-1}{(1+r)-1}=\frac{a}{r}\left[(1+r)^n-1\right] \]

Step 4: Compare Y with X.
Multiply X by \((1+r)^n\):
\[ (1+r)^n\cdot X=(1+r)^n\cdot\frac{a}{r}\left[1-(1+r)^{-n}\right]=\frac{a}{r}\left[(1+r)^n-1\right] \]
This is exactly Y, so \(Y=X(1+r)^n\).

Step 5: Verify with a quick numeric check.
Let \(a=1\), \(r=1\), \(n=2\). Then \(X=\frac{1}{2}+\frac{1}{4}=0.75\), and \(Y=1+2=3\).
Check: \(X(1+r)^n=0.75\times2^2=0.75\times4=3\), which matches Y exactly.

Step 6: Rule out the other options.
Using the same numbers: option (A) gives \(0.75\times(2+4)=4.5\), option (C) gives \(0.75\times\frac{2^2-1}{1}=2.25\), and option (D) gives \(0.75\times2=1.5\). None of these equal 3, so only \(X(1+r)^n\) is correct.

Final Answer:
\[ Y=X(1+r)^n \]
\[ \boxed{X(1+r)^n} \]