Generate a clear step-by-step proof of the formula for the sum of the first \(n\) natural numbers using the classic pairing method.
The sum of the first \(n\) positive integers is one of the fundamental formulas in arithmetic and discrete mathematics.
In sigma notation, the same relationship is written as
The formula lets you calculate the sum directly without individually adding all \(n\) numbers.
The first and last terms always add to \(n+1\):
The next pair has the same total:
The next pair also has the same total:
This pattern continues through the entire sequence. Therefore, when the forward and backward sums are added, every resulting pair equals \(n+1\).
There are exactly \(n\) such pairs, which gives \(n(n+1)\) for \(2S\). Dividing by two produces the final formula.
Let \(n=10\). The sum is
Reverse the sequence:
Adding them term-by-term gives:
There are ten copies of 11:
Therefore, \(1+2+3+\cdots+10=55\).
The sequence \(1,2,3,\ldots,n\) is an arithmetic sequence with first term \(1\) and common difference \(1\).
The general arithmetic-series formula is
For the natural-number sequence:
Substituting these values gives
The number \(\frac{n(n+1)}{2}\) is also called the \(n\)-th triangular number.
Triangular numbers represent quantities that can be arranged in a triangular pattern. The first few are:
Each triangular number is therefore the sum of the first \(n\) positive integers.
The proof does not depend on choosing a particular numerical value for \(n\). The variable \(n\) represents any positive integer.
Because the first and last terms always pair to \(n+1\), every corresponding pair has the same total. Since the sequence contains \(n\) terms, adding the forward and reversed sums creates \(n\) identical pairs.
This is a general algebraic proof rather than merely a numerical example.
Explore other Summe tools for sums, sequences, arithmetic patterns, and related mathematical formulas:
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The formula is \[ \sum_{k=1}^{n}k = \frac{n(n+1)}{2}. \]
The pairing proof writes the sum forward and backward, then adds corresponding terms. Every pair equals \(n+1\), producing \(n(n+1)\) for twice the sum. Dividing by two gives \(n(n+1)/2\).
A commonly told historical account describes Carl Friedrich Gauss using a pairing idea for an elementary arithmetic-sum problem during his school years. The pairing argument itself is a general arithmetic technique and does not depend on the historical story.
In the convention used by this calculator, the first 10 positive natural numbers are \(1,2,3,\ldots,10\), and their sum is \(55\).
Yes. It is the arithmetic-series formula specialized to the sequence \(1,2,3,\ldots,n\).
It is \[ \sum_{k=1}^{n}k. \] This compact notation means to add every integer from 1 through \(n\).
The \(n\)-th triangular number is \[ T_n=\frac{n(n+1)}{2}. \] It is exactly the sum of the first \(n\) positive integers.
Yes. The variable \(n\) represents an arbitrary positive integer, so the algebraic pairing argument establishes the formula for every positive integer \(n\).