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To solve the given summation:

\[
\sum_{i=1}^{10} 3 \times 2^{i-1}
\]

We can factor out the constant \( 3 \) from the summation:

\[
3 \times \sum_{i=1}^{10} 2^{i-1}
\]

This expression is now a geometric series, where the first term \(a = 2^0 = 1\) and the common ratio \(r = 2\). The number of terms \(n = 10\).

The sum \(S_n\) of the first \(n\) terms of a geometric series is given by:

\[
S_n = a \frac{r^n - 1}{r - 1}
\]

Substituting the values:

\[
S_{10} = 1 \times \frac{2^{10} - 1}{2 - 1} = 2^{10} - 1
\]

Calculate \(2^{10}\):

\[
2^{10} = 1024
\]

So,

\[
S_{10} = 1024 - 1 = 1023
\]

Thus, the sum of the series is:

\[
3 \times 1023 = 3069
\]

So, the exact answer is \( \boxed{3069} \).

Would you like more details or have any questions?

Here are five related questions to consider:
1. How would the result change if the summation limit was from \( i = 0 \) to \( i = 9 \)?
2. What if the common ratio was \( r = 3 \) instead of \( r = 2 \)?
3. How do you derive the formula for the sum of a geometric series?
4. What is the effect of multiplying the series by a constant, as done in this problem?
5. How would you solve this problem if the exponent was \( i+1 \) instead of \( i-1 \)?

**Tip:** When dealing with geometric series, always identify the first term and common ratio before applying the summation formula.

Evaluate the summation ∑ (3 × 2^(i-1)) from i=1 to 10. Give an exact answer, do not round.

Learn how to evaluate the summation of the series 3 * 2^(i-1) from i=1 to 10 with an exact answer. This problem covers key algebraic concepts such as geometric series, suitable for students in Grades 9-12.

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