prove that summation m=0 of n 3powern =3powern+1 "-1/2"using mathematical induction 

We are tasked with proving the formula for the sum of powers of 3 using **mathematical induction**:

\[
S(n) = \sum_{m=0}^{n} 3^m = \frac{3^{n+1} - 1}{2}
\]

### Steps of Mathematical Induction

#### 1. **Base Case:**

First, let's check if the formula holds true for \( n = 0 \).

The left-hand side (LHS) is:

\[
S(0) = \sum_{m=0}^{0} 3^m = 3^0 = 1
\]

The right-hand side (RHS) is:

\[
\frac{3^{0+1} - 1}{2} = \frac{3^1 - 1}{2} = \frac{3 - 1}{2} = \frac{2}{2} = 1
\]

Since both sides are equal, the base case holds true.

#### 2. **Inductive Hypothesis:**

Assume that the formula holds true for some \( n = k \). That is:

\[
\sum_{m=0}^{k} 3^m = \frac{3^{k+1} - 1}{2}
\]

We need to prove that the formula also holds true for \( n = k + 1 \). In other words, we need to show:

\[
\sum_{m=0}^{k+1} 3^m = \frac{3^{(k+1)+1} - 1}{2}
\]

#### 3. **Inductive Step:**

Using the inductive hypothesis for \( n = k \), we can express the sum up to \( k+1 \) as:

\[
\sum_{m=0}^{k+1} 3^m = \left( \sum_{m=0}^{k} 3^m \right) + 3^{k+1}
\]

By the inductive hypothesis, we know:

\[
\sum_{m=0}^{k} 3^m = \frac{3^{k+1} - 1}{2}
\]

Therefore, substituting this into the equation gives:

\[
\sum_{m=0}^{k+1} 3^m = \frac{3^{k+1} - 1}{2} + 3^{k+1}
\]

Next, we need to simplify the right-hand side:

\[
= \frac{3^{k+1} - 1}{2} + \frac{2 \cdot 3^{k+1}}{2}
= \frac{3^{k+1} - 1 + 2 \cdot 3^{k+1}}{2}
= \frac{3^{k+1} + 2 \cdot 3^{k+1} - 1}{2}
= \frac{3 \cdot 3^{k+1} - 1}{2}
= \frac{3^{k+2} - 1}{2}
\]

This is exactly what we needed to prove for \( n = k+1 \).

#### 4. **Conclusion:**

Since the base case holds true and the inductive step is valid, by mathematical induction, the formula:

\[
\sum_{m=0}^{n} 3^m = \frac{3^{n+1} - 1}{2}
\]

is true for all \( n \geq 0 \).

---

Would you like further details or clarifications on any step?

### Related Questions:
1. How do we apply mathematical induction to other types of summation problems?
2. Can the same approach be used for proving formulas involving geometric series?
3. What are the common pitfalls when performing mathematical induction?
4. How does the choice of base case affect the induction proof?
5. Can we generalize this proof for sums of other powers, like \( 4^m \)?

### Tip:
Always ensure that your inductive hypothesis clearly matches the format of the formula you are trying to prove, and check each step of simplification carefully!

prove that summation m=0 of n 3powern =3powern+1 -1/2 using mathematical induction

This detailed guide demonstrates how to prove the formula for the sum of powers of 3, S(n) = sum(m=0 to n) 3^m = (3^(n+1) - 1) / 2, using mathematical induction. Suitable for students in Grades 10-12, it covers the steps of induction, including the base case and inductive step.

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