# 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