Sigma notation, indices and changed limits
| English | Français |
|---|---|
| summation index/sʌˈmeɪʃn ˈɪndeks/ | summation index |
A sum starts at the third term. How many terms does it contain, and what changes if we rename or shift its index?
- A sum starts at the third term. How many terms does it contain, and what changes if we rename or shift its index?
- This lesson studies summation index 求和指标: The variable that takes each integer value between a sum’s stated limits.
Choose the mathematical structure
- In a sum from k=a to b, substitute each integer a,a+1,…,b into the summand and add the results. There are b−a+1 terms when a≤b. The summation index is a dummy variable, but other parameters keep their values. Renaming an index preserves the sum only when its bound occurrences are renamed consistently.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines summation index?
The variable that takes each integer value between a sum’s stated limits.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
The sum from k=3 to 7 of (2k−1) is 5+7+9+11+13=45, with five terms. It also equals the sum from k=1 to 7 minus the sum from k=1 to 2: 49−4=45. With j=k−2, the lower limit becomes 1, the upper becomes 5 and 2k−1 becomes 2j+3, again giving 45. By linearity, the original sum is 2 times the sum of k from 3 to 7, minus five copies of 1: 2×25−5=45.
Sigma notation, indices and changed limits
In a sum from k=a to b, substitute each integer a,a+1,…,b into the summand and add the results
Check whether a sequence or sum argument preserves all its defining information.
Evaluate the sum of (2k−1) from k=3 to 7.
Add 5+7+9+11+13=45.
Test a tempting shortcut
- An inclusive sum from 3 to 7 has five terms, not four or seven. Subtract the prefix through a−1, not through a, when changing the lower limit. A constant is added once per index value: the sum of c from a to b is c(b−a+1). Shifting the index changes both bounds and the summand.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
Renaming or shifting an index never requires changing the summand. This claim is false. Explain which definition or assumption it violates.
How many terms occur in that sum?
Count inclusive integer indices: 7−3+1=5.
Renaming or shifting an index never requires changing the summand.
An inclusive sum from 3 to 7 has five terms, not four or seven. Subtract the prefix through a−1, not through a, when changing the lower limit. A constant is added once per index value: the sum of c from a to b is c(b−a+1). Shifting the index changes both bounds and the summand.
Interpret a new situation
- For the sum of (2k−1) from k=1 to n, the index k changes while n is the fixed upper bound; the result is n² for positive integer n. Thus a tail from k=m to n is n²−(m−1)² for 1≤m≤n. Expand a few terms to check a symbolic manipulation before applying a sum formula.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
After j=k−2, what is the new upper limit?
When k=7, j=7−2=5.
Match each part of a complete solution to its purpose.
An assumption justifies the model; a check tests the result; interpretation connects it to the question.
Use this in your course
- 7357 · A-level · D. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
The variable that takes each integer value between a sum’s stated limits. Choose the relationship, show the method, check its assumptions and interpret the result.