Telescoping series
Find $S_r = 1^r + 2^r + 3^r + ... + n^r$ where r is any fixed positive integer in terms of $S_1, S_2, ... S_{r-1}$.
Problem
In 1654 Blaise Pascal published a general method for summing powers of positive integers, i.e. summing all the series. $$S_1 = 1 + 2 + 3 + ... + n$$ $$S_2 = 1^2 + 2^2 + 3^2 + ... + n^2$$ $$S_3 = 1^3 + 2^3 + 3^3 + ... + n^3$$ $$\dots$$ $$S_r = 1^r + 2^r + 3^r + ... + n^r$$Pascal's method uses the coefficients which appear in Pascal's triangle and in the Binomial Theorem,
first finding $S_1$, and then using $S_1$ to find $S_2$, and then using both to find $S_3$, and so on. The method applies, where $r$ is any fixed positive integer, to: $$S_r =\sum_{k=1}^n k^r.$$
Case 1: $r = 1$
Simplify $(k+1)^2 - k^2$ and hence show that: $$ \sum_{k=1}^n [(k+1)^2-k^2] = {2\choose 1}S_1 +n. $$ Write the sum out in full for $n = 6$. Use this method to prove that $S_1=n(n+1)/2$.
Case 2: $r = 2$
Simplify $(k+1)^3 - k^3$ and hence show that $$ \sum_{k=1}^n[(k+1)^3-k^3] = (n+1)^3-1 = {3\choose 1}S_2 + {3\choose 2}S_1 + n. $$ Hence prove that $S_2 = n(n+1)(2n + 1)/6$.
Case 3: $r = 3$
Use this technique to find $1^3 + 2^3 + 3^3 + ... + 10^3$.
General case
Show that $$ (n+1)^r - (n+1) = {r\choose 1}S_1 + {r\choose 2}S_2 + {r\choose 3}S_3 + \dots {r\choose r-1}S_{r-1}. $$
Getting Started
The question gives you instructions step by step.
Student Solutions
Congratulations to Herbert Pang of Sha Tin College, Hong Kong and also to Ka Wing Kerwin Hui for their excellent solutions. Both of these solutions were written in Word 97 using Equation Editor 3.0 and are beautifully presented.
Case 1: $r=1$
We simplify $(k + 1)^2 - k^2$. Writing the binomial coefficients in the form $$ {n \choose r} = \frac{n!}{r!(n-r)!}, $$ then for each $k$, $$ (k + 1)^2 - k^2 = k^2 + {2\choose 1}k + 1 - k^2 = {2\choose 1}k + 1;$$ hence $$\sum_{k=1}^n[(k + 1)^2-k^2] =\sum_{k=1}^n\left[{2\choose 1}k + 1\right] =\sum_{k=1}^n{2\choose 1}k + \sum_{k=1}^n 1.$$ Writing $S_r$ for $\sum^n_{k=1} k^r$, this gives $$\sum_{k=1}^n[(k + 1)^2-k^2] ={2\choose 1}S_1 + n.$$ Writing this sum in full for $n=6$ we note that terms cancel out in pairs (hence the name 'telescoping series') giving: $$[2^2-1^2] + [3^2-2^2] + [4^2-3^2] + [5^2-4^2] + 6^2-5^2] + [7^2-6^2]= -1 + 49 = 48$$ If we write this out in full with a general $n$ we get $$[2^2-1^2] + [3^2-2^2] + [4^2-3^2] + \cdots + [(n + 1)^2-n^2] = -1 + (n + 1)^2 = n^2 + 2n.$$ Hence $$ n^2 + 2n = 2S_1 + n,$$ and this gives $$ S_1 = n(n + 1)/2.$$
Case 2: $r=2$
We simplify $(k + 1)^3 - k^3$. For any $k$, $$ (k + 1)^3 - k^3 = k^3 + {3\choose 1}k^2 + {3\choose 2}k + 1 - k^3 = {3\choose 1}k^2 + {3\choose 2}k + 1,$$ and adding this for $k=1,\ldots , n$, we get $$ \sum_{k=1}^n[(k + 1)^3-k^3] =\sum_{k=1}^n\left[{3\choose 1}k^2 + {3\choose 2}k + 1\right] ={3\choose 1}S_2 + {3\choose 2}S_1 + n.$$ The left hand side is a telescoping series, and is $$ [2^3-1^3] + [3^3-2^3] + [4^3-3^3] + \cdots + [(n + 1)^3-n^3] = -1 + (n + 1)^3 = n^3 + 3n^2 + 3n,$$ so that $$ n^3 + 3n^2 + 3n = {3\choose 1}S_2 + {3\choose 2}S_1 + n.$$ As we have already found $S_1$ to be $n(n + 1)/2$, we can now find the formula for $S_2$: $$ n^3 + 3n^2 + 3n = 3S_2 + 3n(n + 1)/2 + n.$$ Simplifying this gives $$ S_2 = {n(n + 1)(2n + 1)\over 6}.$$
Case 3: $r=3$
We simplify $(k + 1)^4 - k^4$. For any $k$, $$(k + 1)^4 - k^4 = k^4 + {4\choose 1}k^3 + {4\choose 2}k^2 + {4\choose 3}k + 1 - k^4 = {4\choose 1}k^3 + {4\choose 2}k^2 + {4\choose 3}k + 1$$and adding this for $k=1,\ldots , n$, we get $$\sum_{k=1}^n[(k + 1)^4-k^4] = \sum_{k=1}^n\left[{4\choose 1}k^3 + {4\choose 2}k^2 + {4\choose 3}k + 1\right] = {4\choose 1}S_3 + {4\choose 2}S_2 + {4\choose 3}S_1 + n$$The left hand side is a telescoping series, and is $$ [2^4-1^4] + [3^4-2^4] + [4^4-3^4] + \cdots + [(n + 1)^4-n^4] = -1 + (n + 1)^4 = n^4 + 4n^3 + 6n^2 + 4n$$ so that $$ n^4 + 4n^3 + 6n^2 + 4n = {4\choose 1}S_3 + {4\choose 2}S_2 + {4\choose 3}S_1 + n$$ Using the formulae for $S_1$ and $S_2$, we obtain an equation for $S_3$, and simplifying this we get $$ S_3 = {n^2(n + 1)^2\over 4}.$$ For $n=10$ we get $$ \sum_{k=1}^{10}k^3 = {10^2\times 11^2\over 4} = 3025.$$ It is interesting to note that for all $n$, $$S_3 = \left({n(n + 1)\over 2}\right)^2 = \big(S_2\big)^2.$$
General case
We simplify $(k + 1)^r - k^r$. For any $k$, $$(k + 1)^r - k^r = {r\choose 1}k^{r-1} + {r\choose 2}k^{r-2} + \cdots + {r\choose r-1}k + 1,$$ and adding this for $k=1,\ldots , n$, we get $$ \sum_{k=1}^n[(k + 1)^r-k^r] = {r\choose 1}S_{r-1} + {r\choose 2}S_{r-2} + \cdots + {r\choose r-1}S_1 + n$$ The left hand side is a telescoping series, and is $$ [2^r-1^r] + [3^r-2^r] + [4^r-3^r] + \cdots + [(n + 1)^r-n^r] = -1 + (n + 1)^r$$ hence $$ (n + 1)^r-1 = {r\choose 1}S_{r-1} + {r\choose 2}S_{r-2} + \cdots + {r\choose r-1}S_1 + n$$ Transferring $n$ from the right to the left, and using $$ {r\choose k} = {r\choose r-k}$$ we get $$(n + 1)^r-(n + 1) = {r\choose 1}S_{1} + {r\choose 2}S_{2} + \cdots + {r\choose r-1}S_{r-1}$$
Teachers' Resources
Why do this problem?
The problem gives step by step guidance so that learners only need to apply what they know about the Binomial expansion of $(k+1)^n$ and do some simple algebraic manipulation to be able to find general formulae for the sums of powers of the integers. As the name suggests the method makes use of the 'telescoping' property so that all the intermediate terms disappear leaving only the first and last.
Possible
Approach
You might choose to introduce this method just for the sum of
the squares of the integers as pictured in the pyramid
illustration.
Key question
What is $[2^2-1^2] + [3^2-2^2] +
[4^2-3^2] + \cdots + [(n + 1)^2-n^2]$?
Possible support
Try the problems
Natural Sum,
More Sequences and Series, and
OK Now Prove It
Possible support
Read the article
Proof by Induction.
Possible extension
Seriesly