Changing Indices in Summation- Techniques and Examples

What Is Index Changing in Summation Notation?

When you work with summation notation, you'll often need to change the index variable or its range. This isn't some advanced trick—it's a basic manipulation skill you'll use constantly in calculus, series analysis, and discrete math.

Index changing means rewriting a sum so that it starts at a different number, uses a different variable, or covers the same terms in a different form. The goal is always the same: express the same sum differently without changing what you're actually adding up.

Why Bother Changing Indices?

You need to change indices when:

• The index range doesn't match what you need for an operation
• You're combining two sums and need matching indices
• A formula or theorem requires a specific starting value
• You're comparing two series and need identical forms

In short: mathematical operations require compatible notation. You can't add sums with different index ranges directly—you have to make them match first.

The Core Technique: Index Shifting

The most common operation is shifting—moving the starting index while adjusting the expression accordingly.

The Basic Formula

If you have a sum starting at i = m and want it to start at i = k, use this relationship:

i = j + (m - k)

Or more practically: let j = i - (m - k)

The new index j starts where you want it, and you substitute back into the original expression.

Simple Example

Change this sum to start at j = 1:

∑(i=3 to 10) (2i + 1)

Here's the process:

• Original starts at i = 3. You want j = 1.
• The shift amount is 3 - 1 = 2
• So j = i - 2, which means i = j + 2
• Substitute: 2(j + 2) + 1 = 2j + 5
• New sum: ∑(j=1 to 8) (2j + 5)

Count the terms: 10 - 3 + 1 = 8 terms. The new upper limit is 1 + 8 - 1 = 8. ✓

Working With Different Starting Points

Let's do more examples to cement the pattern.

Example 1: Shifting Down

Rewrite ∑(k=0 to 5) (k² + 3k) so it starts at k = 1.

You want the new index to start at 1, and the original starts at 0. The shift is 1.

• Let j = k + 1, so k = j - 1
• Substitute: (j - 1)² + 3(j - 1) = j² - 2j + 1 + 3j - 3 = j² + j - 2
• New upper limit: 5 + 1 = 6
• Result: ∑(j=1 to 6) (j² + j - 2)

Example 2: Shifting to Start at Zero

Rewrite ∑(n=5 to 12) 3ⁿ with an index starting at 0.

Original starts at 5. You want 0. Shift amount: 5.

• Let j = n - 5, so n = j + 5
• Substitute: 3^(j+5) = 3^5 · 3^j = 243 · 3^j
• New range: j = 0 to 7 (since 12 - 5 = 7)
• Result: 243 · ∑(j=0 to 7) 3^j

Notice the constant factor 243 came out front. That's allowed—pull out anything that doesn't depend on the index.

Reversing the Order of Summation

Sometimes you need to flip a double sum or swap the order of indices. This comes up constantly in multivariable calculus.

The Key Insight

For a double sum ∑(i=a to b) ∑(j=c to d) f(i,j), the bounds matter. If j's limits depend on i, you can't just swap blindly.

Independent bounds: You can swap freely.

Dependent bounds: You need to redetermine the new limits carefully.

Example: Independent Bounds

Evaluate ∑(i=1 to 3) ∑(j=4 to 5) (i + j)

Since bounds don't depend on each other:

• Inner sum first: for each i, sum j from 4 to 5
• i + 4 + i + 5 = 2i + 9
• Outer sum: ∑(i=1 to 3) (2i + 9) = (2 + 9) + (4 + 9) + (6 + 9) = 11 + 13 + 15 = 39

Swapping order gives the same answer: ∑(j=4 to 5) ∑(i=1 to 3) (i + j) = 39.

Combining and Splitting Sums

You often need to combine multiple sums or split one into parts.

Combining Sums

Two sums with the same index range can be combined:

∑aᵢ + ∑bᵢ = ∑(aᵢ + bᵢ)

If ranges differ, expand to matching indices first.

Example

Simplify: ∑(i=1 to n) i + ∑(i=1 to n) 2

Second sum is just adding 2, n times: 2n

First sum is n(n+1)/2

Combined: n(n+1)/2 + 2n = (n² + n + 4n)/2 = (n² + 5n)/2

Splitting Sums

You can split a sum at any point:

∑(i=m to p) aᵢ = ∑(i=m to k) aᵢ + ∑(i=k+1 to p) aᵢ

Where m ≤ k < p.

Common Mistakes to Avoid

These errors show up constantly. Don't make them.

How To: Step-by-Step Process

Here's your checklist for any index change problem:

1. Identify the current index and bounds (i = m to n)
2. Determine the target starting index (k)
3. Calculate the shift: shift = m - k
4. Define the new index: j = i - shift, so i = j + shift
5. Substitute into the expression being summed
6. Calculate the new upper bound: n - shift
7. Verify the number of terms matches

Quick Reference Table

Practice Problems

Try these without looking at the answers:

1. Rewrite ∑(i=2 to 9) (3i - 4) with an index starting at 1.
2. Change ∑(k=1 to 5) (k² + k) to start at k = 0.
3. Simplify ∑(i=1 to n) (i + 1) - ∑(i=1 to n) i

Answers

Problem 1: Shift = 2 - 1 = 1. New index j = i - 1. Expression: 3(j+1) - 4 = 3j - 1. Upper bound: 9 - 1 = 8. Result: ∑(j=1 to 8) (3j - 1)

Problem 2: Shift = 1 - 0 = 1. New index j = k - 1. Expression: (j+1)² + (j+1) = j² + 2j + 1 + j + 1 = j² + 3j + 2. Upper bound: 5 - 1 = 4. Result: ∑(j=0 to 4) (j² + 3j + 2)

Problem 3: The second sum is ∑i. The first is ∑i + ∑1 = ∑i + n. Subtracting gives n. The answer is n.

When You'll Actually Use This

Index changing shows up in:

• Series convergence tests — comparison tests often require matching index starting points
• Changing integration bounds — similar logic applies when switching variables
• Algorithm analysis — when summing over array indices with shifted ranges
• Probability distributions — reindexing pmfs and expected value calculations

If you're taking discrete math, calculus series units, or anything involving summations, these skills aren't optional. They're the foundation.