WorkbenchEach term 1/(k(k+1)) has been split into a difference of two numbers. Press “Cancel out!” and see which numbers the slanted lines join. Then drag n up. Does more survive?
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What this move does
The general terman of a sequence {an} and the sum of the first n termsSn=a1+a2+⋯+an determine each other:
a1=S1,an=Sn−Sn−1(n≥2).
Going the other way, from the general term to the sum, one of the most useful ideas is telescoping: if every term can be written as the difference of one function at two places,
ak=c(V(k)−V(k+g)),
then when you add, the −V(k+g) in term k cancels exactly with the +V(k+g) in term k+g. The steps:
Split the general term into “this minus next”, and multiply back to check the constant c.
Write out the first g rows and the last g rows, and see clearly what cancels.
Only the first g “this” parts and the last g “next” parts survive.
The slanted lines in the figure at the top are step 2: each line joins a pair of numbers that cancel, g rows apart.
Predict first
1⋅21+2⋅31+⋯+99⋅1001=? (fractions are allowed)
Choose an answer first; then you can see the result.
✓You predicted it!✗Not what you expected — which is exactly why it is worth a pause.
k(k+1)1=k1−k+11. In the 99 rows, everything cancels in pairs except the 11 in the first row and the −1001 in the last:
1−1001=10099.
In general k=1∑nk(k+1)1=n+1n. Drag n in the figure at the top to check.
Work through one
Example: find 1/(1·3) + 1/(2·4) + … + 1/(n(n+2))
Step 1 / 4
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Split first
The two factors in the denominator differ by 2. Try k1−k+21=k(k+2)2: there is an extra 2, so we must multiply by 21 in front. The “=” on the left of each row in the picture says the split is right.
k(k+2)1=21(k1−k+21)
Write it out
Cancel
Any n
Not only fractions
Anything that can be written as “this minus next” telescopes. k⋅k!=(k+1)⋅k!−k!=(k+1)!−k!, so
1⋅1!+2⋅2!+⋯+n⋅n!=(n+1)!−1.
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This time the “next” number is the bigger one, so we write (k+1)!−k!. Press “Cancel out!”: what remains is (n+1)! and 1!.
The product of three consecutive integers works the same way: k(k+1)(k+2)1=21(k(k+1)1−(k+1)(k+2)1).
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1⋅2⋅31+⋯+5⋅6⋅71=21(21−421)=215.
When it fails
Break it
Counterexample lab: remove one condition
Conclusion: What survives at the ends is the sum← still true?
k(k+3)1=31(k1−k+31), with constant 31. After cancelling, the two readouts agree.
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Forget the constant and write k(k+3)1=k1−k+31.
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The left of every row now shows “=”. The cancelling still goes through, and the survivors can still be added up, but they give the value of a different sum: exactly 3 times the correct answer. Telescoping itself never complains; only multiplying back catches the mistake.
Practice
Exercise 1Basic
Find 1⋅21+2⋅31+⋯+20⋅211.
Hint · direction
Each term can be written as a difference of two unit fractions.
Hint · method
k(k+1)1=k1−k+11: the end of one row cancels the start of the next.
Hint · first step
What survives is the 1 in the first row and the −211 in the last.
Full solution
1−211=2120.
Exercise 2Basic
The sum of the first n terms of a sequence is Sn=2n2−n. Find the general term an (in terms of n).
Hint · direction
For which n does an=Sn−Sn−1 hold?
Hint · method
Use the difference for n≥2 first, then compute a1=S1 separately and see whether the two merge.
Hint · first step
Sn−1=2(n−1)2−(n−1)=2n2−5n+3.
Full solution
For n≥2, an=(2n2−n)−(2n2−5n+3)=4n−3; and a1=S1=1=4⋅1−3 fits too. So an=4n−3.
Exercise 3Advanced
Find k=1∑10k(k+2)1.
Hint · direction
It is the same sum as in “Work through one”, with n=10.
Hint · method
Sn=21(1+21−n+11−n+21).
Hint · first step
n=10: what survives is 1,21,−111,−121.
Full solution
21(23−111−121)=21⋅132198−12−11=264175.
Exercise 4Challenge
Find 1+21+2+31+⋯+99+1001.
Hint · direction
These are not differences of fractions, but they can become differences of square roots.
Hint · method
Rationalize the denominator: multiply by k+1−k.
Hint · first step
k+k+11=(k+1)−kk+1−k=k+1−k.
Full solution
When you add, everything in the middle cancels, leaving 100−1=10−1=9.
Using Sₙ − Sₙ₋₁ for the general term and forgetting n = 1
✗
The wrong step: Given Sn=n2+2n+1, we have Sn=(n+1)2, so an=Sn−Sn−1=(n+1)2−n2=2n+1.
What went wrong? Think first, then open
Diagnosisan=Sn−Sn−1 only makes sense for n≥2 (S0 is not defined). a1 must be found separately: a1=S1=4, while the formula 2n+1 gives 3 at n=1. Check: if an=2n+1 held for every n, then S1 would be 3, not 4.
✓
The right step: a1=S1=4, and for n≥2, an=Sn−Sn−1=2n+1. So an={4,2n+1,n=1n≥2. The two pieces merge into one formula only when the constant term of Sn is 0.
“Always check a₁ = S₁ separately”
General Terms and Sums ← → step · F full screen · Esc exit