Math Problem Statement

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

Solution

The problem provided is the summation:

i=1n2i\sum_{i=1}^{n} 2i

This represents the sum of the sequence 2i2i from i=1i=1 to i=ni=n. Let's break this down and solve it step by step.

General formula:

The summation can be factored as:

i=1n2i=2i=1ni\sum_{i=1}^{n} 2i = 2 \sum_{i=1}^{n} i

We know that the sum of the first nn natural numbers is given by the formula:

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

Substituting this back into the equation:

2i=1ni=2×n(n+1)2=n(n+1)2 \sum_{i=1}^{n} i = 2 \times \frac{n(n+1)}{2} = n(n+1)

Thus, the summation simplifies to:

i=1n2i=n(n+1)\sum_{i=1}^{n} 2i = n(n+1)

Final answer:

The result of the summation i=1n2i\sum_{i=1}^{n} 2i is n(n+1)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 nn?
  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 ii?

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

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Math Problem Analysis

Mathematical Concepts

Summation
Algebra

Formulas

Sum of first n natural numbers: \(\sum_{i=1}^{n} i = \frac{n(n+1)}{2}\)
Summation: \(\sum_{i=1}^{n} 2i = n(n+1)\)

Theorems

Summation of arithmetic series

Suitable Grade Level

Grades 9-12