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The problem provided is the summation:

\[
\sum_{i=1}^{n} 2i
\]

This represents the sum of the sequence \(2i\) from \(i=1\) to \(i=n\). Let's break this down and solve it step by step.

### General formula:
The summation can be factored as:

\[
\sum_{i=1}^{n} 2i = 2 \sum_{i=1}^{n} i
\]

We know that the sum of the first \(n\) natural numbers is given by the formula:

\[
\sum_{i=1}^{n} i = \frac{n(n+1)}{2}
\]

### Substituting this back into the equation:

\[
2 \sum_{i=1}^{n} i = 2 \times \frac{n(n+1)}{2} = n(n+1)
\]

Thus, the summation simplifies to:

\[
\sum_{i=1}^{n} 2i = n(n+1)
\]

### Final answer:

The result of the summation \(\sum_{i=1}^{n} 2i\) is \(n(n+1)\).

Would you like more details on any part of this? Here are some related questions:

1. What is the general formula for summing natural numbers?
2. How would the result change if the summation limits changed?
3. How can we apply this formula to specific values of \(n\)?
4. What happens if we sum an arithmetic progression with a different common difference?
5. Can this formula be generalized to summations involving higher powers of \(i\)?

**Tip**: Always try to break summations into known formulas to simplify calculations.

Evaluate the summation: \(\sum_{i=1}^{n} 2i\)

Learn how to evaluate the summation \(\sum_{i=1}^{n} 2i\), which simplifies to \(n(n+1)\). This algebra problem is ideal for students in grades 9-12 and involves using the formula for the sum of the first n natural numbers.

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