# Sum of n, n², or n³

The sum of positive integers \(n\), the sum of their squares \(n^2\), and the sum of their cubes \(n^3\) are all series that can each be reduced down to a simple formula. The derivation of this formula is a critical component in algebra, particularly in the study of series, because it allows students to simplify or break apart series.

At first, the derivation of these sums may seem to have limited use. However manipulations of them yield useful results for topics like string theory, quantum mechanics, and complex numbers. In more advanced mathematics, summations of positive integers and exponents of positive integers are frequently used to to reduce down divergent series.

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## Sum of first \(n\) positive integers

Let \(S_n = 1+2+3+4+\cdots +n = \displaystyle \sum_{k=1}^n k.\) Since this is an arithmetic series with initial term \(a=1,\) common difference \(d=1,\) and last term \(l = n,\) \[S_n = \frac{n}{2}(1+n)=\frac{n(n+1)}{2}.\]

An alternative way of proving is by rearranging the equation as follows: \[\begin{eqnarray} S_n & = & 1 & + & 2 & + & 3 & + \ldots + & n \\ S_n & = & n & + & n-1 & + & n-2 & + \ldots + & 1 .\\ \end{eqnarray}\] Grouping and adding the above two sums, we get \[\begin{eqnarray} 2S_n & = & (1+n)+(2+n-1)+(3+n-2) + \cdots + (n+1) \\ & = & \underbrace{(n+1)+(n+1)+(n+1)+\cdots+(n+1)}_{n\ \text{times}} \\ & = & n(n+1). \end{eqnarray}\] Thus, \[S_n = \dfrac{n(n+1)}{2}.\]

Note: You can also prove this by induction .

## Find the sum of the first \(100\) positive integers.

Plugging \(n=100\) in our equation, \[1+2+3+4+\dots + 100 = \frac{100(101)}{2} = \frac{10100}{2},\] which implies our final answer is 5050. \( _\square \)

## Find the sum of the first \( 100 \) even positive integers.

The first \( 100 \) even natural numbers are \( 2, 4, 6, 8, \ldots, 198, 200 \), and we want to find the sum \( 2 + 4 + 6 + \cdots + 198 + 200\). Note that each term is divisible by 2. We can take a factor of two common from each term; which will transform the summation into a form that we are familiar with:

\[ 2 + 4 + 6 + \cdots + 198 + 200 = 2 \times ( 1 + 2 + 3 + \cdots + 99 + 100 ) .\]

We know how to find the sum of the first 100 natural numbers:

\[ 1 + 2 + 3 + \cdots + 99 + 100 = \frac{ 100 \times 101 } {2} = 5050.\]

Thus, \( 2 \times ( 1 + 2 + 3 + \cdots + 99 + 100) = 2 \times 5050 = 10100 .\) \(_\square\)

## Sum of the squares of first \(n\) positive integers

Consider the identity \[(x+1)^3-x^3=3x^2+3x+1.\] Substituting \(x = 1, 2, 3, \dots , (n-1), n\) successively, we get \[\begin{align} 2^3-1^3&=3\cdot 1^2 + 3\cdot 1 +1\\ 3^3-2^3&=3\cdot 2^2 + 3\cdot 2 +1\\ 4^3-3^3&=3\cdot 3^2 + 3\cdot 3 +1\\ &\vdots \\ n^3-(n-1)^3&=3\cdot (n-1)^2 + 3\cdot (n-1) +1\\ (n+1)^3-n^3&=3\cdot n^2 + 3\cdot n +1. \end{align}\] Adding column-wise, we obtain \[\begin{align} (n+1)^3 -1^3&= 3 \sum_{k=1}^n k^2 + 3 \sum_{k=1}^n k +n\\ \Rightarrow n^3+3n^2+3n &= 3 \sum_{k=1}^n k^2 + 3\frac{n(n+1)}{2} +n. \end{align}\] Simplifying, we get \[3 \displaystyle \sum_{k=1}^n k^2 = \frac{2n^3+3n^2+n}{2}.\] Hence, \[\displaystyle \sum_{k=1}^n k^2 = 1^2+2^2+3^2+ \dots + n^2 = \frac {n(n+1)(2n+1)}{6}. \]

## Find the sum of the squares of first \(100\) positive integers.

Plugging \(n=100\) in our equation, \[1^2+2^2+3^2+4^2+\dots + 100^2 = \frac{100(101)(201)}{6} = \frac{2030100}{6},\] which implies our final answer is 338350. \( _\square \)

## Sum of the cubes of first \(n\) positive integers

Consider the identity \[(x+1)^4-x^4=4x^3+6x^2+4x+1.\] Subsituting \(x = 1, 2, 3, \dots , (n-1), n\) successively, we get \[\begin{align} 2^4-1^4&=4\cdot 1^3 + 6\cdot 1^2 +4\cdot 1+1\\ 3^4-2^4&=4\cdot 2^3 + 6\cdot 2^2 +4\cdot 2+1\\ 4^4-3^4&=4\cdot 3^3 + 6\cdot 3^2 +4\cdot 3+1\\ &\vdots \\ n^4-(n-1)^4&=4\cdot (n-1)^3 + 6\cdot (n-1)^2 +4\cdot (n-1)+1\\ (n+1)^4-n^4&=4\cdot n^3 + 6\cdot n^2 +4\cdot n+1. \end{align}\] Adding column-wise, we obtain \[\begin{align} (n+1)^4 -1^4&= 4 \displaystyle \sum_{k=1}^n k^3 + 6 \displaystyle \sum_{k=1}^n k^2 + 4 \displaystyle \sum_{k=1}^n k+n\\ \Rightarrow n^4+4n^3+6n^2+4n &= 4 \displaystyle \sum_{k=1}^n k^3 + 6\frac{n(n+1)(2n+1)}{6} +4 \frac{n(n+1)}{2} +n. \end{align}\] Simplifying, we get \[ 4 \displaystyle \sum_{k=1}^n k^3 = n^4+2n^3+n^2.\] Hence, \[\displaystyle \sum_{k=1}^n k^3 = 1^3+2^3+3^3+\cdots+n^3 = \frac{n^2(n+1)^2}{4}.\]

## Find the sum of the cubes of first \(100\) positive integers.

Plugging \(n=100\) in our equation, \[1^3+2^3+3^3+4^3+\dots + 100^3 = \frac{100^2(101^2)}{4} = \frac{102010000}{4},\] which implies our final answer is 25502500. \( _\square \)

Now try this problem:

**Cite as:**Sum of n, n², or n³.

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