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: Write out X term by term.
\[ X=\frac{a}{1+r}+\frac{a}{(1+r)^2}+\cdots+\frac{a}{(1+r)^n} \]
We want $Y=a+a(1+r)+a(1+r)^2+\cdots+a(1+r)^{n-1}$, and notice both series use the same building block $a(1+r)^k$, just with different (and opposite-signed) powers of $(1+r)$.

Step 2: Multiply every term of X by $(1+r)^n$.
Multiplying each term of $X$ by $(1+r)^n$ shifts each power of $(1+r)$ up by $n$:
\[ (1+r)^n\cdot X=(1+r)^n\cdot\frac{a}{1+r}+(1+r)^n\cdot\frac{a}{(1+r)^2}+\cdots+(1+r)^n\cdot\frac{a}{(1+r)^n} \]
\[ =a(1+r)^{n-1}+a(1+r)^{n-2}+\cdots+a(1+r)^{1}+a(1+r)^{0} \]

Step 3: Recognize this as Y, just written in reverse order.
The terms $a(1+r)^{n-1}+a(1+r)^{n-2}+\cdots+a(1+r)+a$ are exactly the same terms as $Y=a+a(1+r)+\cdots+a(1+r)^{n-1}$, only listed from the highest power down to the lowest instead of the lowest up to the highest.
Addition does not care about the order of the terms, so this sum equals $Y$.

Step 4: State the result.
\[ (1+r)^n\cdot X=Y \]
So $Y=X(1+r)^n$, obtained here purely by shifting powers, without ever writing down the closed-form geometric series formula.

Step 5: Quick check.
With $a=1,r=1,n=2$: $X=0.75$, and $(1+r)^n\cdot X=4\times0.75=3$, matching $Y=1+2=3$ directly.

Final Answer:
$Y=X(1+r)^n$, option (B).
$\boxed{X(1+r)^n}$