Math Problem Statement

What is the summation of (-1)^(i+1) * i^2 for i=1 to n?

Solution

The image shows the following summation:

i=1n(1)i+1i2\sum_{i=1}^{n} (-1)^{i+1} \cdot i^2

This is a finite series where each term alternates in sign based on (1)i+1(-1)^{i+1}, and each term is the square of ii. Let's break it down:

  • The term (1)i+1(-1)^{i+1} ensures alternating signs.
  • The term i2i^2 is simply the square of the index ii.

General Form:

For i=1i = 1 to nn: S=(1)212+(1)322+(1)432++(1)n+1n2S = (-1)^{2} \cdot 1^2 + (-1)^{3} \cdot 2^2 + (-1)^{4} \cdot 3^2 + \dots + (-1)^{n+1} \cdot n^2 This series alternates between positive and negative squares of integers.

Do you want further details on how to compute or analyze this summation for a specific nn? Let me know!

Here are some related questions:

  1. What is the closed form of this summation for specific nn?
  2. How can we generalize this series to non-integer exponents?
  3. What are the properties of alternating series, and when do they converge?
  4. How can this be related to alternating power series expansions?
  5. What real-world problems might this summation model?

Tip: Alternating series can often be simplified by grouping terms based on symmetry or cancellation patterns.

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Math Problem Analysis

Mathematical Concepts

Summation
Alternating Series
Powers

Formulas

S = sum_{i=1}^{n} (-1)^(i+1) * i^2

Theorems

Alternating Series Test
Power Series Expansion

Suitable Grade Level

Grades 10-12