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Accumulation from Sums · Tutorial 927 of 1495

Recovering Totals from Successive Changes

Derive the telescoping sum for successive changes and use it to check cumulative data and motion calculations.

Beginner 9 min read

What You'll Learn

  • Define the change in a quantity between consecutive observations
  • Derive why successive changes sum to the endpoint difference
  • Check a cumulative table by reconstructing its values step by step
  • Recover a missing change from the starting value, ending value, and other changes
  • Distinguish net displacement from total distance in piecewise-constant motion

From Successive Changes to a Net Change

A quantity may be recorded at a sequence of times or stages. The entries in such a record can show the quantity itself, such as a tank’s volume, or the change in the quantity from one observation to the next. These two kinds of information are linked: adding the successive signed changes recovers the overall change from the first observation to the last.

This link is useful for checking cumulative tables. If a table begins with one value and reports changes at each step, the final value must agree with the sum of those changes. In motion, the same idea says that adding signed displacements over consecutive time intervals gives the net displacement over the whole time period.

Definition: Let \(Q_0,Q_1,\ldots,Q_n\) be the values of a quantity at consecutive observations. The change during step \(i\), for \(i=1,\ldots,n\), is \(\Delta Q_i=Q_i-Q_{i-1}\). A positive change means the quantity increased; a negative change means it decreased.

The word signed matters. An increase contributes positively and a decrease contributes negatively. The sum of the changes gives the net change, not necessarily the total amount of increase or decrease considered separately.

The Telescoping Sum

Write each change as the value after the step minus the value before it. When these expressions are added, every intermediate value appears once with a positive sign and once with a negative sign. Only the last and first values remain.

$$ \begin{aligned} \sum_{i=1}^{n}\Delta Q_i &=\sum_{i=1}^{n}(Q_i-Q_{i-1})\\ &=(Q_1-Q_0)+(Q_2-Q_1)+\cdots+(Q_n-Q_{n-1})\\ &=Q_n-Q_0. \end{aligned} $$

This cancellation pattern is called telescoping: the intermediate terms cancel like the overlapping sections of a collapsing telescope. The identity does not require the changes to be equal, or all to have the same sign. It only requires that the change at each step be calculated from the values at the two consecutive endpoints.

Theorem: For any finite sequence of values \(Q_0,Q_1,\ldots,Q_n\), if \(\Delta Q_i=Q_i-Q_{i-1}\), then \(\sum_{i=1}^{n}\Delta Q_i=Q_n-Q_0\). The sum of successive changes equals the final value minus the initial value.

The equality can also be read in the opposite direction. If the initial value and all the changes are known, the final value is the initial value plus the sum of the changes. Each change updates the current value, so the updates can be carried out one at a time or checked altogether against the endpoint difference.

Worked Example: Checking a Cumulative Table

Worked Example: Checking a Cumulative Table

A tank contains \(120\) liters at the start. During three successive periods, its reported volume changes are \(+27\) liters, \(-8\) liters, and \(+21\) liters. Check the final entry \(160\) liters against the changes.

First reconstruct the volume after each period:

$$ \begin{aligned} Q_0&=120\text{ L},\\ Q_1&=Q_0+27=120+27=147\text{ L},\\ Q_2&=Q_1-8=147-8=139\text{ L},\\ Q_3&=Q_2+21=139+21=160\text{ L}. \end{aligned} $$

Now add the reported changes, keeping their signs:

$$ 27+(-8)+21=27-8+21=19+21=40\text{ L}. $$

The endpoint difference is \(160-120=40\) liters, so the sum matches the endpoint difference. The table is consistent: the reconstructed final volume is \(160\) liters, as reported.

Check: The total change of \(40\) liters added to the starting volume gives \(120+40=160\) liters, the stated final value.

A useful audit therefore has two forms. Reconstructing the table checks each individual step; comparing the total of the changes with the endpoint difference checks the overall result. Agreement in the second check does not by itself prove that every intermediate entry is correct, because errors in different rows might offset one another. When the intermediate data matter, check each step as well.

Using the Identity to Find a Missing Change

If one change in a sequence is unknown, the endpoint difference determines what the total of all changes must be. Subtract the known changes from that required total to find the missing one. This works because the unknown change must make the whole sum equal the endpoint difference.

Worked Example: Recovering a Missing Change

A quantity starts at \(35\) units and ends at \(58\) units. Four successive changes are \(+12\), \(-5\), an unknown change \(d\), and \(+9\) units. Find \(d\).

The required total change is the endpoint difference:

$$ 58-35=23\text{ units}. $$

The known changes sum to \(12+(-5)+9=12-5+9=7+9=16\) units. Therefore \(d\) must satisfy \(16+d=23\), giving \(d=23-16=7\) units.

Check the complete sum: \(12+(-5)+7+9=12-5+7+9=7+7+9=14+9=23\) units. Starting from \(35\), the successive values are \(35+12=47\), \(47-5=42\), \(42+7=49\), and \(49+9=58\). The reconstructed endpoint agrees with the stated value.

Successive Changes in Motion

For motion along a line, position plays the role of \(Q\). The displacement over an interval is the ending position minus the starting position. Thus, for consecutive time intervals, the signed displacements add to the overall displacement. A negative displacement indicates motion in the negative direction; it is not a negative distance traveled.

When velocity is constant on an interval, the displacement on that interval is velocity multiplied by elapsed time. If the velocity has different constant values on consecutive intervals, calculate each signed displacement separately, then add them. This is an application of successive changes: each interval changes the position from its value at the beginning to its value at the end.

Worked Example: Net Displacement and Distance

An object starts at position \(10\) meters. Its velocity is \(3\) meters per second from \(t=0\) to \(t=2\) seconds, \(-2\) meters per second from \(t=2\) to \(t=5\) seconds, and \(4\) meters per second from \(t=5\) to \(t=7\) seconds. Find its final position, net displacement, and total distance traveled.

On the first interval, the duration is \(2-0=2\) seconds, so the displacement is \(3(2)=6\) meters. The position at \(t=2\) is \(10+6=16\) meters.

On the second interval, the duration is \(5-2=3\) seconds. Its signed displacement is \((-2)(3)=-6\) meters. The position at \(t=5\) is \(16+(-6)=10\) meters.

On the third interval, the duration is \(7-5=2\) seconds. Its displacement is \(4(2)=8\) meters. The final position is \(10+8=18\) meters.

The sum of the signed displacements is \(6+(-6)+8=6-6+8=0+8=8\) meters. The endpoint difference checks this result: \(18-10=8\) meters. The total distance is instead \(6+6+8=20\) meters, because distance counts the length of each part of the motion as nonnegative.

Check: Starting at \(10\) meters and adding the net displacement gives \(10+8=18\) meters, the final position found from the intervals. The object traveled \(6+6+8=20\) meters in total.

What the Sum Does—and Does Not—Tell You

The telescoping identity compresses a sequence of changes into one overall change. It tells you the difference between the final and initial values, even if the quantity rose and fell many times in between. It does not tell you the largest or smallest intermediate value, nor how much total increase occurred separately from total decrease.

For motion, this distinction is especially important. Net displacement depends on direction and can be zero even when an object has traveled a positive distance. Total distance is found by adding the lengths traveled on each interval; net displacement is found by adding signed displacements, or simply by subtracting the initial position from the final position.

Takeaway: Add successive changes with their signs to find net change. For motion, this gives displacement. To find total distance, account for the length of travel on each interval rather than canceling motion in opposite directions.

A Common Sign Mistake

A frequent error is to replace a negative change by its positive magnitude when finding net change. In the motion example, writing \(6+6+8=20\) meters as the net displacement is wrong: it has counted the backward displacement as positive. The correct signed calculation is \(6+(-6)+8=8\) meters. The value \(20\) meters is the total distance, not the net displacement.

The reverse mistake is also possible: canceling opposite-direction changes when the question asks for distance. Read the quantity requested before adding. Net change uses signed changes; total distance uses nonnegative lengths.

Check Your Understanding

Use signed changes where appropriate. For each question, check your result against the relevant endpoint or definition.

  1. A quantity starts at \(18\) units and changes by \(+7\), \(-11\), and \(+4\) units. What is its final value, and what is its net change?
  2. A table lists starting and ending values of \(42\) and \(67\). Three reported changes are \(+10\), \(-3\), and \(d\). Find \(d\) and verify the total change.
  3. An object moves \(5\) meters in the positive direction and then \(9\) meters in the negative direction. What are its net displacement and total distance?
  4. Why can the sum of successive changes be checked using only the first and last values?
  5. Does knowing the net change determine the total amount of increase? Explain briefly.

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