Answer
1. \(\displaystyle 1 + 2 + 2^2 + \dots + 2^{n-1} \;=\; 2^{\,n}-1.\)
2. \(\displaystyle 1 + 3 + 3^2 + \dots + 3^{n-1} \;=\; \frac{3^{\,n}-1}{2}.\)
3. \(\displaystyle 2 + 2^2 + \dots + 2^n \;=\; 2^{\,n+1} - 2.\)
4. \(\displaystyle \sum_{k=0}^n (-1)^k\,2^k
\;=\;
\frac{1 - (-2)^{\,n+1}}{3}.\)
All of these follow directly from the geometric‐series identity
\[
1 + r + r^2 + \cdots + r^n
\;=\;
\frac{r^{\,n+1}-1}{r-1}
\quad (r \neq 1).
\]
Work Step by Step
\[
1 \;+\; r \;+\; r^2 \;+\;\cdots\;+\;r^n
\;=\;
\frac{r^{\,n+1} - 1}{r - 1}
\quad
(\text{valid for }r\neq1).
\]
---
## (a) A Sum of Powers of 2
**Problem:** For an integer \(n \ge 1\), find a closed‐form expression for
\[
1 \;+\; 2 \;+\; 2^2 \;+\;\cdots\;+\;2^{n-1}.
\]
This is just a geometric sum with common ratio \(r = 2\), starting at \(2^0=1\) and going up to \(2^{n-1}\). Using the formula:
\[
1 + 2 + 2^2 + \dots + 2^{n-1}
\;=\;
\sum_{k=0}^{n-1} 2^k
\;=\;
\frac{2^{\,n} - 1}{2 - 1}
\;=\;
2^{\,n} \;-\; 1.
\]
---
## (b) A Sum of Powers of 3
**Problem:** For an integer \(n \ge 1\), find a closed‐form expression for
\[
1 \;+\; 3 \;+\; 3^2 \;+\;\cdots\;+\;3^{n-1}.
\]
Here the common ratio is \(r = 3\), and we sum from \(3^0=1\) up to \(3^{n-1}\). Thus,
\[
1 + 3 + 3^2 + \dots + 3^{n-1}
\;=\;
\sum_{k=0}^{n-1} 3^k
\;=\;
\frac{3^{\,n} - 1}{3 - 1}
\;=\;
\frac{3^{\,n} - 1}{2}.
\]
---
## (c) Another Sum of Powers of 2 (Starting at \(2^1\))
**Problem:** For an integer \(n \ge 2\), find a closed‐form expression for
\[
2 \;+\; 2^2 \;+\; 2^3 \;+\;\cdots\;+\;2^n.
\]
Note that this is \(\sum_{k=1}^n 2^k\). We can write
\[
\sum_{k=1}^n 2^k
\;=\;
\bigl(\sum_{k=0}^n 2^k\bigr) \;-\; 2^0
\;=\;
\bigl(2^{\,n+1} - 1\bigr) \;-\; 1
\;=\;
2^{\,n+1} \;-\; 2.
\]
Hence,
\[
2 + 2^2 + \cdots + 2^n
\;=\;
2^{\,n+1} - 2.
\]
*(If instead you needed \(1 + 2 + 2^2 + \dots + 2^n\), that was part (a).)*
---
## (d) An Alternating Sum of Powers of 2
**Problem (typical form):** For an integer \(n \ge 1\), find a closed‐form for
\[
2^0 \;-\; 2^1 \;+\; 2^2 \;-\; 2^3 \;+\;\cdots\;+\;(-1)^n\,2^n.
\]
Observe this is a geometric series with first term \(a = 1\) (that is \(2^0\)) and common ratio \(r = -2\). Indeed,
\[
\sum_{k=0}^n (-1)^k\,2^k
\;=\;
\sum_{k=0}^n \bigl((-2)\bigr)^k.
\]
By the geometric‐series formula,
\[
\sum_{k=0}^n \bigl((-2)\bigr)^k
\;=\;
\frac{(-2)^{\,n+1} - 1}{(-2) - 1}
\;=\;
\frac{(-2)^{\,n+1} - 1}{-3}
\;=\;
\frac{1 - (-2)^{\,n+1}}{3}.
\]
Thus,
\[
2^0 \;-\; 2^1 \;+\; 2^2 \;-\; 2^3 \;+\;\cdots\;+\;(-1)^n\,2^n
\;=\;
\frac{1 \;-\; (-2)^{\,n+1}}{3}.
\]
