Certainly! We'll use the principle of mathematical induction to prove that for all natural numbers [tex]\( n \)[/tex]:
[tex]\[
\sum_{i=1}^n 3^i = \frac{3(3^n - 1)}{2}
\][/tex]
### Step 1: Base Case
First, we prove the base case where [tex]\( n = 1 \)[/tex].
For [tex]\( n = 1 \)[/tex]:
[tex]\[
\sum_{i=1}^1 3^i = 3^1 = 3
\][/tex]
If we substitute [tex]\( n = 1 \)[/tex] in the right-hand side (RHS) of our original equation:
[tex]\[
\frac{3(3^1 - 1)}{2} = \frac{3(3 - 1)}{2} = \frac{3 \cdot 2}{2} = 3
\][/tex]
Both sides are equal when [tex]\( n = 1 \)[/tex], so the base case holds.
### Step 2: Induction Hypothesis
Suppose the statement is true for some integer [tex]\( k \geq 1 \)[/tex]. That is, assume
[tex]\[
\sum_{i=1}^k 3^i = \frac{3(3^k - 1)}{2}
\][/tex]
is true.
### Step 3: Induction Step
We need to prove that the statement is true for [tex]\( n = k + 1 \)[/tex]. That is, we need to prove:
[tex]\[
\sum_{i=1}^{k+1} 3^i = \frac{3(3^{k+1} - 1)}{2}
\][/tex]
Consider the left-hand side (LHS) for [tex]\( n = k + 1 \)[/tex]:
[tex]\[
\sum_{i=1}^{k+1} 3^i = \left( \sum_{i=1}^k 3^i \right) + 3^{k+1}
\][/tex]
Using the induction hypothesis [tex]\(\sum_{i=1}^k 3^i = \frac{3(3^k - 1)}{2}\)[/tex]:
[tex]\[
\sum_{i=1}^{k+1} 3^i = \frac{3(3^k - 1)}{2} + 3^{k+1}
\][/tex]
Now, we need to combine these terms into the form of the right-hand side (RHS):
[tex]\[
\frac{3(3^k - 1)}{2} + 3^{k+1} = \frac{3(3^k - 1) + 2 \cdot 3^{k+1}}{2}
\][/tex]
Simplify the numerator:
[tex]\[
3(3^k - 1) + 2 \cdot 3^{k+1} = 3 \cdot 3^k - 3 + 2 \cdot 3^{k+1}
\][/tex]
Notice that [tex]\(2 \cdot 3^{k+1}\)[/tex] can be written as [tex]\(2 \cdot 3 \cdot 3^k\)[/tex]:
[tex]\[
3 \cdot 3^k - 3 + 2 \cdot 3 \cdot 3^k = 3 \cdot 3^k - 3 + 6 \cdot 3^k
\][/tex]
[tex]\[
= 9 \cdot 3^k - 3 = 3^{k+1} \cdot 3 - 3 = 3^{k+1}(3) - 3
\][/tex]
[tex]\[
= 3(3^{k+1} - 1)
\][/tex]
Therefore:
[tex]\[
\frac{3(3^k - 1)}{2} + 3^{k+1} = \frac{3(3^{k+1} - 1)}{2}
\][/tex]
So, we have shown that if the formula holds for [tex]\( n = k \)[/tex], it also holds for [tex]\( n = k + 1 \)[/tex].
### Conclusion
Since the base case and the induction step have been proved, by the principle of mathematical induction, the statement is true for all natural numbers [tex]\( n \)[/tex]:
[tex]\[
\sum_{i=1}^n 3^i = \frac{3(3^n - 1)}{2}
\][/tex]
