The Reverse Order Addition Method

The reverse order addition method is the clever technique that young Gauss used to compute $1+2+\cdots+100$. This method exploits the symmetry of a sequence, simplifying the computation by adding the sequence in forward and reverse order.

The Principle of the Method

Basic steps:

1. Write the forward sum: $S_n = a_1 + a_2 + a_3 + \cdots + a_n$
2. Write the reverse sum: $S_n = a_n + a_{n-1} + a_{n-2} + \cdots + a_1$
3. Add the two equations: $2S_n = (a_1 + a_n) + (a_2 + a_{n-1}) + \cdots + (a_n + a_1)$
4. If every pair of sums is equal, then $2S_n = n(a_1 + a_n)$

Applicable Scenarios

The reverse order addition method applies to sequences satisfying the following condition:

Typical examples:

• Arithmetic sequences: $a_k + a_{n-k+1} = a_1 + a_n$
• Binomial coefficients: $C_n^k + C_n^{n-k} = C_n^k + C_n^k$ (when $k = n-k$)

A Classic Application: Gaussian Summation

The method young Gauss used to compute $1+2+3+\cdots+100$:

$$\begin{aligned} S &= 1 + 2 + 3 + \cdots + 98 + 99 + 100 \\ S &= 100 + 99 + 98 + \cdots + 3 + 2 + 1 \\ \hline 2S &= 101 + 101 + 101 + \cdots + 101 + 101 + 101 \quad (\text{100 copies of 101}) \end{aligned}$$

$2S = 100 \times 101 = 10100$

$S = 5050$

This is exactly the derivation of the arithmetic sequence summation formula!

Worked Examples

Example 1: Summing an Arithmetic Sequence

Find the sum: $S_n = 1 + 3 + 5 + \cdots + (2n-1)$

Solution:

$$\begin{aligned} S_n &= 1 + 3 + 5 + \cdots + (2n-1) \\ S_n &= (2n-1) + (2n-3) + (2n-5) + \cdots + 1 \\ \hline 2S_n &= 2n + 2n + 2n + \cdots + 2n \quad (\text{$n$ copies of $2n$}) \end{aligned}$$

$2S_n = n \times 2n = 2n^2$

$S_n = n^2$

Example 2: A Symmetric Sequence

Find the sum: $S_n = \frac{1}{1+\sqrt{n}} + \frac{1}{1+\sqrt{n-1}} + \cdots + \frac{1}{1+\sqrt{2}} + \frac{1}{1+\sqrt{1}}$

Solution:

Let $a_k = \frac{1}{1+\sqrt{k}}$. Then:

$S_n = a_n + a_{n-1} + \cdots + a_2 + a_1$

In reverse order:

$S_n = a_1 + a_2 + \cdots + a_{n-1} + a_n$

Observe: is $a_k + a_{n-k+1}$ a constant?

Actually, this example does not satisfy the simple symmetry condition and requires other methods.

Example 3: Summing Binomial Coefficients

Find the sum: $S_n = C_n^0 + C_n^1 + C_n^2 + \cdots + C_n^n$

Solution:

Using the binomial coefficient property $C_n^k = C_n^{n-k}$:

$$\begin{aligned} S_n &= C_n^0 + C_n^1 + C_n^2 + \cdots + C_n^n \\ S_n &= C_n^n + C_n^{n-1} + C_n^{n-2} + \cdots + C_n^0 \\ \hline 2S_n &= (C_n^0 + C_n^n) + (C_n^1 + C_n^{n-1}) + \cdots \end{aligned}$$

But a simpler method is to use the binomial theorem: $(1+1)^n = 2^n$

Practice Problems

Exercise 1

Find the sum: $S_n = 2 + 4 + 6 + \cdots + 2n$

Exercise 2

Find the sum: $S = 1 \times 2024 + 2 \times 2023 + 3 \times 2022 + \cdots + 2024 \times 1$

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Summary

Symbols Used in This Article