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.