Every closed-form count this course derives — a distribution total, a generating function's coefficient, a recurrence's solution — eventually comes down to rewriting a finite sum into a shape a known formula can close. Three moves do almost all of that work: shifting the index, telescoping, and swapping the order of a double sum.
Index shift is just renaming. The sum 1^2+2^2+\cdots+n^2 doesn't care whether the running label starts at 1 or 0. The same numbers get added either way. Telescoping is a row of dominoes falling into each other. Each term's arrival cancels the previous term's departure. Only the very first departure and the very last arrival survive. The same domino trick works on a chain of multiplications instead of additions. A running product collapses the same way a running sum does. It cancels everything but its very first and very last piece.
Index shift: replacing the summation variable (j = i - c, say) relabels which term each position in the sum names, without changing which numbers get added — \sum_{i=1}^{n} i^2 = \sum_{j=0}^{n-1}(j+1)^2 is the same sum written two ways. Telescoping: when a sum's terms are each a difference of consecutive values of some function, almost every term cancels against its neighbor, leaving only the first and last: \sum_{i=1}^{n}\bigl(f(i)-f(i+1)\bigr) = f(1)-f(n+1). Swapping order: a double sum \sum_i \sum_j can be reorganized as \sum_j \sum_i whenever the index ranges are handled correctly — genuinely useful once Chapter 9's generating functions stack sums inside sums. Manipulating sums means rewriting them into one of these shapes rather than adding term by term. A finite product is manipulated the same way a sum is: telescoping a ratio of consecutive terms cancels a whole chain down to its first numerator and its last denominator, $\prod_{i=2}^{n}\left(1
- \frac{1}{i}\right) = \frac{1}{n}$ being the cleanest example — the exact move behind the product form of Euler's totient function in Chapter 4.
Shift the index without changing the sum
Pick the rule that relabels the sum without changing which numbers get added.
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Worked example
A telescoping sum collapses to two terms
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1
Evaluate \sum_{i=1}^{n}\left(\dfrac{1}{i}-\dfrac{1}{i+1}\right).
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2
Write out the first few terms: \left(1-\tfrac12\right) + \left(\tfrac12-\tfrac13\right) + \left(\tfrac13-\tfrac14\right) + \cdots. Every middle term is the negative of the next term's first piece.
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3
Every interior term cancels, leaving only the very first "+1" and the very last "-\frac{1}{n+1}."
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4\boxed{1 - \dfrac{1}{n+1}}
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Worked example
Splitting a sum into two closable pieces
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1
Find a closed form for \sum_{i=1}^{n}(2i-1) — the sum of the first n odd numbers.
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2
Split the sum into two pieces the closed forms from Section 8 can each close: 2\sum_{i=1}^n i - \sum_{i=1}^n 1.
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3
Apply \sum_{i=1}^n i = \frac{n(n+1)}{2} and \sum_{i=1}^n 1 = n, then simplify: 2\cdot\frac{n(n+1)}{2} - n = n(n+1) - n.
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4\boxed{n^2}
The sum of the first n odd numbers is always a perfect square — a fact Chapter 6 revisits with a picture instead of algebra.
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Where this goes next
Generating functions
Chapter 9's whole machinery runs on exactly this toolkit: extracting the coefficient of x^n in a product of two power series is a double sum that only becomes tractable after an index shift lines up matching powers, and a telescoping argument is how several of that chapter's closed-form generating functions get derived from their recurrences in the first place.
Check your understanding (counts toward your progress)
Which technique relabels a sum's index without changing which numbers get added?
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Why
An index shift renames the running variable (j = i - c, for instance); the same terms are still added, just under a new label.
A double sum sum_i sum_j a_{ij} is swapped to sum_j sum_i a_{ij}. When is this valid?
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Why
Swapping the summation order is valid whenever the same set of (i,j) pairs is still covered -- the ranges must be re-expressed correctly, but the values summed don't change.
Think about what swapping summation order actually changes: the set of (i,j) pairs being added, or just the order they get listed in.
As long as the new index ranges still describe exactly the same set of pairs, does relisting them in a different order change the total?
A sum's terms are f(1)-f(2), f(2)-f(3), f(3)-f(4). What technique applies, and what's the result?
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Why
Consecutive terms cancel their shared pieces (the -f(2) cancels the next +f(2), and so on), leaving only f(1) - f(4) -- telescoping.
Line up the three terms and check whether any piece of one term cancels a piece of a neighboring term.
Split sum_{i=1}^{n} (4i + 7) into two separate sums added together (using i and n only). Enter the coefficient that multiplies sum_{i=1}^{n} i.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
sum(4i+7) = 4*sum(i) + sum(7) -- the coefficient on the sum-of-i piece is 4.
Correct answer
Also accepted:
Why
Split the sum term by term first: the 4i piece and the 7 piece become two separate sums, each keeping its own coefficient.
Using the split-and-close technique, evaluate sum_{i=1}^{8} (2i - 1).
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Why
This sum of the first 8 odd numbers closes to n^2 = 8^2 = 64, matching split-and-close's result from this section.
Correct answer
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Why
Recognize this as the sum of the first several odd numbers -- a shape this section closes to its own simple formula.
Apply that closed form for the sum of the first n odd numbers directly, using n=8.
Why is index shifting useful before applying a closed-form formula?
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Why
Closed forms are usually memorized starting at a specific index (like i=1 or i=0); shifting lines a sum up with that expected form without changing its value.
A student splits sum_{i=1}^{n} (3i - 5) into 3*sum(i) minus sum(5). What must the second piece be, exactly?
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The constant 5 is added n separate times, once per term -- it isn't just 5 once.
Why
sum_{i=1}^{n} 5 adds the constant 5 exactly n times: 5n, not 5.
A constant added inside a sum still gets added once per index value -- count how many times i runs from 1 to n.
In a telescoping sum sum (f(i) - f(i+1)), which terms survive after cancellation?
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Why
Each term's negative piece cancels the next term's positive piece, leaving only the very first positive piece and the very last negative piece.
Write out the first three terms explicitly and watch which piece of one term cancels against a piece of the next term.
Practice: Rewriting a sum before evaluating it (5 problems · doesn’t affect your progress)
A sum's terms are each a difference, f(i) - f(i+1), and nearly all of them should cancel when added. Which manipulation technique is designed exactly for this shape?
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Index shift relabels the running variable -- it doesn't cause terms to cancel against each other.
Splitting separates a sum into pieces added together -- it doesn't rely on terms canceling.
Swapping order applies to double sums with two indices -- not a single difference-of-consecutive-terms shape.
Why
A sum built from consecutive differences is exactly what telescoping is built to collapse -- nearly every interior piece cancels, leaving only the first and last.
Telescoping
After the substitution j = i - 1, the sum sum_{i=1}^{n} i^2 becomes sum_{j=0}^{n-1} of what expression in j? Enter the full new summand as an expression in j.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Substituting i = j+1 turns i^2 into (j+1)^2, with j running from 0 to n-1.
Correct answer
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Why
A student splits sum_{i=1}^{n} (3i - 5) into 3*sum(i) minus sum(5). What must the second piece be, exactly?
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The constant 5 is added n separate times, once per term -- it isn't just 5 once.
Why
sum_{i=1}^{n} 5 adds the constant 5 exactly n times: 5n, not 5.
A constant added inside a sum still gets added once per index value -- count how many times i runs from 1 to n.
5n
Telescoping collapses sum_{i=2}^{7} (1/i - 1/(i+1)) down to just two surviving pieces. Evaluate it as a decimal, rounded to 4 places.
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Why
Telescoping collapses the sum to 1/2 - 1/8 = 3/8 = 0.375.
Correct answer
Also accepted:
Why
Write out the surviving first and last pieces before dividing -- everything in between cancels.
A sum sum_{i=1}^{n} f(i) is rewritten via the substitution j = i + 2 as sum over j of f(j-2). What are the new bounds on j?
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The bounds must shift along with the index -- they can't stay the same after substituting j = i+2.
Adding 2 to i means j is LARGER than i, not smaller -- the new bounds should shift up, not down.
Only the upper bound shifts up by 2 here -- the lower bound shifts by the same +2, from i=1 to j=3, not staying at 1.
Why
j = i+2 means as i runs from 1 to n, j runs from 1+2=3 to n+2 -- both bounds shift by the same +2 the substitution applies to the index.
Apply the substitution j=i+2 to the ORIGINAL lower and upper bounds on i, not just to the summand.
The lower bound was i=1; add 2 to find where j starts. Do the same for the upper bound.
j runs from 3 to n+2