Question

. Suppose \(A=a_{1}+a_{2}+\cdots+a_{n}\) and \(B=b_{1}+b_{2}+\cdots+\) \(b_{n} .\) Show that \(\sum\left(b_{i}-a_{i}\right)=B-A\), or, that the sum of the differences of the parts is equal to the difference of the sums of the parts.

Short answer

Question: Prove that the sum of the differences of the parts is equal to the difference of the sums of the parts. Answer: To prove this, we showed that \(\sum(b_i - a_i) = B - A\), which means the sum of the differences of the parts is equal to the difference of the sums of the parts.

Step-by-step solution

Calculate the difference of the sums

We are given that \(A = a_1 + a_2 + \cdots + a_n\) and \(B = b_1 + b_2 + \cdots + b_n\). We want to calculate the difference of the sums, which is \(B - A\). So we have: \(B - A = (b_1 + b_2 + \cdots + b_n) - (a_1 + a_2 + \cdots + a_n)\).

Distribute the negative sign

Now we distribute the negative sign from \((-1)(A)\) across the terms inside the parentheses: \(B - A = b_1 + b_2 + \cdots + b_n - a_1 - a_2 - \cdots - a_n\).

Rewrite the expression as a sum

Now we rewrite the expression as a sum of the differences of each corresponding term: \(B - A = (b_1 - a_1) + (b_2 - a_2) + \cdots + (b_n - a_n)\).

Recognize the summation notation

Finally, we recognize that this expression can be written using summation notation as: \(B - A = \sum_{i=1}^n(b_i - a_i)\). Therefore, we have shown that \(\sum(b_i - a_i) = B - A\), which means the sum of the differences of the parts is equal to the difference of the sums of the parts.

Key concepts

Summation Notation

Summation notation is a succinct way to express the addition of a sequence of numbers. It is represented by the Greek letter sigma (\(\Sigma\)), and it signifies the sum of multiple terms in a sequence. This notation simplifies the process of writing long sums explicitly.
When you see something like \(\sum_{i=1}^n a_i\), it tells you to add up all the elements from \(a_1\) to \(a_n\). The symbol beneath the sigma specifies how the indices of the sequence begin and end. It's essentially a shortcut to avoid writing very long addition operations repeatedly.
So, when we express a series of terms like \((b_1 - a_1) + (b_2 - a_2) + \cdots + (b_n - a_n)\) using summation notation, it simplifies to \(\sum_{i=1}^n (b_i - a_i)\). This is extremely useful in proving statements about sums or working with numerous terms.

Algebraic Manipulation

Algebraic manipulation involves systematically using operations to simplify or rearrange expressions. It includes processes like expanding expressions, factoring, and applying distributive properties. This is crucial when working with expressions that contain multiple terms or complex structures.
Consider the step where we found the difference between two sums: \(B - A = (b_1 + b_2 + \cdots + b_n) - (a_1 + a_2 + \cdots + a_n)\). To manipulate algebraically, we distributed the negative sign across the second collection of terms: \(B - A = b_1 + b_2 + \cdots + b_n - a_1 - a_2 - \cdots - a_n\).

• The distributive property allows us to handle subtraction by treating it as addition of negative terms.
• This ensures each term in the second sum is accurately subtracted from its counterpart in the first sum.

This step enables us to clearly see how each individual pair of terms contributes to the overall sum, preparing us to express it as one unified summation statement.

Difference of Sums

The difference of sums can be understood by analyzing two sums and comparing them. In the problem at hand, the difference between two sums \(A\) and \(B\) is key.
The exercise involves showing that subtracting individual elements pairwise and then summing produces the same result as simply taking the difference of the total sums. Mathematically, it means demonstrating \(\sum(b_i - a_i) = B - A\).

• This conveys that slicing through the elements, performing subtraction followed by summation, mirrors a broader comparison of whole sums.
• This equivalency helps us break down larger complex problems into smaller manageable equations.

Understanding this helps in recognizing patterns and solving similar problems where such relationships exist between parts and wholes. It is a demonstration of distributive logic at the scale of sequential elements.