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Sum of n, n², or n³ | Brilliant Math & Science Wiki (1)

The series \(\sum\limits_{k=1}^n k^a = 1^a + 2^a + 3^a + \cdots + n^a\) gives the sum of the \(a^\text{th}\) powers of the first \(n\) positive numbers, where \(a\) and \(n\) are positive integers. Each of these series can be calculated through a closed-form formula. The case \(a=1,n=100\) is famously said to have been solved by Gauss as a young schoolboy: given the tedious task of adding the first \(100\) positive integers, Gauss quickly used a formula to calculate the sum of \(5050.\)

The formulas for the first few values of \(a\) are as follows:

\[\begin{align}\sum_{k=1}^n k &= \frac{n(n+1)}2 \\\sum_{k=1}^n k^2 &= \frac{n(n+1)(2n+1)}6 \\\sum_{k=1}^n k^3 &= \frac{n^2(n+1)^2}4.\end{align}\]

Faulhaber's formula, which is derived below, provides a generalized formula to compute these sums for any value of \(a.\)

Manipulations of these sums yield useful results in areas including string theory, quantum mechanics, and complex numbers.

Contents

  • Sum of the First \(n\) Positive Integers
  • Sum of the Squares of the First \(n\) Positive Integers
  • Sum of the Cubes of the First \(n\) Positive Integers
  • Some Examples
  • Generalizations
  • Faulhaber's Formula
  • See Also

Sum of the First \(n\) Positive Integers

Let \(S_n = 1+2+3+4+\cdots +n = \displaystyle \sum_{k=1}^n k.\) The elementary trick for solving this equation (which Gauss is supposed to have used as a child) is a rearrangement of the sum as follows:

Sum of n, n², or n³ | Brilliant Math & Science Wiki (2)

\[\begin{eqnarray}S_n & = & 1 & + & 2 & + & 3 & + \cdots + & n \\S_n & = & n & + & n-1 & + & n-2 & + \cdots + & 1 .\\\end{eqnarray}\]

Grouping and adding the above two sums gives

\[\begin{eqnarray}2S_n & = & (1+n)+(2+n-1)+(3+n-2) + \cdots + (n+1) \\ & = & \underbrace{(n+1)+(n+1)+(n+1)+\cdots+(n+1)}_{n\ \text{times}} \\& = & n(n+1).\end{eqnarray}\]

Therefore,

\[S_n = \dfrac{n(n+1)}{2}.\]

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. \( _\square \)

Show that the sum of the first \(n\) positive odd integers is \(n^2.\)

There are several ways to solve this problem. One way is to view the sum as the sum of the first \(2n\) integers minus the sum of the first \(n\) even integers. The sum of the first \(n\) even integers is \(2\) times the sum of the first \(n\) integers, so putting this all together gives

\[\frac{2n(2n+1)}2 - 2\left( \frac{n(n+1)}2 \right) = n(2n+1)-n(n+1) = n^2.\]

Even more succinctly, the sum can be written as

\[\sum_{k=1}^n (2k-1) = 2\sum_{k=1}^n k - \sum_{k=1}^n 1 = 2\frac{n(n+1)}2 - n = n^2.\ _\square\]

In a similar vein to the previous exercise, here is another way of deriving the formula for the sum of the first \(n\) positive integers. Start with the binomial expansion of \((k-1)^2:\)

\[(k-1)^2 = k^2 - 2k + 1.\]

Rearrange the terms as below:

\[k^2-(k-1)^2 = 2k-1.\]

Now sum both sides:

\[\sum_{k=1}^n \big(k^2-(k-1)^2\big) = 2 \sum_{k=1}^n k - \sum_{k=1}^n 1.\]

The left sum telescopes: it equals \(n^2.\) The right side equals \(2S_n - n,\) which gives \(2S_n - n = n^2,\) so \(S_n = \frac{n(n+1)}2.\)

This technique generalizes to a computation of any particular power sum one might wish to compute.

Sum of the Squares of the First \(n\) Positive Integers

Continuing the idea from the previous section, start with the binomial expansion of \((k-1)^3:\)

\[(k-1)^3 = k^3 - 3k^2 + 3k - 1.\]

Rearrange the terms:

\[k^3-(k-1)^3=3k^2-3k+1.\]

As before, summing the left side from \(k=1\) to \(n\) yields \(n^3.\) This gives

\[\begin{align}n^3 &= 3 \left( \sum_{k=1}^n k^2 \right) - 3 \sum_{k=1}^n k + \sum_{k=1}^n 1 \\n^3 &= 3 \left( \sum_{k=1}^n k^2 \right) - 3 \frac{n(n+1)}2 + n \\3 \left( \sum_{k=1}^n k^2 \right) &= n^3 + 3 \frac{n(n+1)}2 - n \\\Rightarrow \sum_{k=1}^n k^2 &= \frac13 n^3 + \frac12 n^2 + \frac16 n \\&= \frac{n(n+1)(2n+1)}6.\end{align}\]

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

Plugging in \(n=100,\)

\[1^2+2^2+3^2+4^2+\dots + 100^2 = \frac{100(101)(201)}{6} = \frac{2030100}{6} = 338350.\ _\square\]

Sum of the Cubes of the First \(n\) Positive Integers

Again, start with the binomial expansion of \((k-1)^4\) and rearrange the terms:

\[k^4-(k-1)^4=4k^3-6k^2+4k-1.\]

Sum from \(1\) to \(n\) to get

\[n^4 = 4 s_{3,n} - 6 s_{2,n} + 4 s_{1,n} - n.\]

Here \(s_{a,n}\) is the sum of the first \(n\) \(a^\text{th}\) powers. So

\[\begin{align}4s_{3,n} &= n^4 + 6 \frac{n(n+1)(2n+1)}6 - 4 \frac{n(n+1)}2 + n \\\\s_{3,n} &= \frac14 n^4 + \frac12 n^3 + \frac34 n^2 + \frac14 n - \frac12 n^2 - \frac12 n + \frac14 n \\\\s_{3,n} &= \frac14 n^4 + \frac12 n^3 + \frac14 n^2 \\\\&= \frac{n^2(n+1)^2}4.\end{align}\]

Find the sum of the cubes of the first \(200\) positive integers.

Plugging \(n=200\) in our equation,\[1^3+2^3+3^3+4^3+ 5^3 + 6^3 + 7^3 +8^3 \dots + 200^3 = \frac{200^2\big(201^2\big)}{4} = \frac{1616040000}{4} = 404010000.\ _\square\]

\[\large \displaystyle\sum_{n=1}^{10}n\big(1+n+n^2\big)= \, ? \]

Some Examples

Simplify

\[2 + 4 + 6 + \cdots + 2n.\]

We have

\[\begin{align}2+4+6+\cdots+2n&=\sum _{ i=1 }^{ n }{ 2i } \\ &=2(1+2+3+\cdots+n)\\ &=2\times \frac { n(n+1) }{ 2 } \\&=n(n+1).\ _\square\end{align}\]

Simplify

\[1+3+5+\cdots+(2n-1).\]

We have

\[\begin{align}1+3+5+\cdots+(2n-1)&=\sum _{ i=1 }^{ n }{ (2i-1) } \\&=\sum _{ i=1 }^{ n }{ 2i } -\sum _{ i=1 }^{ n }{ 1 } \\ &=2\sum _{ i=1 }^{ n }{ i } -n\\ &=2\times \frac { n(n+1) }{ 2 } -n\\ &=n(n+1)-n\\ &=n(n+1-1)\\ &={ n }^{ 2 }.\ _\square\end{align}\]

Simplify

\[2^2+4^2+6^2+\cdots+(2n)^2.\]

We have

\[\begin{align}2^2+4^2+6^2+\cdots+(2n)^2&=\sum_{i=1}^{n}(2i)^2\\&=\sum_{i=1}^{n}\big(2^2 i^2\big)\\ &=4\sum _{ i=1 }^{ n }{ { i }^{ 2 } } \\ &=4\cdot \frac { n(n+1)(2n+1) }{ 6 } \\ &=\frac { 2n(n+1)(2n+1) }{ 3 }.\ _\square\end{align} \]

Simplify

\[1^2+3^2+5^2+\cdots+(2n-1)^2.\]

We have

\[\begin{align}1^2+3^2+5^2+\cdots+(2n-1)^2&=\left(1^2+2^2+3^2+4^2+\cdots+(2n-1)^2+(2n)^2\right)-\left(2^2+4^2+6^2+\cdots+(2n)^2\right)\\&=\sum_{i=1}^{2n} i^2-\sum_{i=1}^{n}(2i)^2\\&=\frac{2n(2n+1)(4n+1)}{6}-\frac{2n(n+1)(2n+1)}{3}\\&=\frac{n(2n+1)\big((4n+1)-2(n+1)\big)}{3}\\&=\frac{n(2n-1)(2n+1)}{3}.\ _\square\end{align}\]

Generalizations

As in the previous section, let \(s_{a,n} = \sum\limits_{k=1}^n k^a.\) Then the relevant identity, derived in the same way from the binomial expansion, is

\[n^{a+1} = \binom{a+1}1 s_{a,n} - \binom{a+1}2 s_{a-1,n} + \binom{a+1}3 s_{a-2,n} - \cdots + (-1)^{a-1} \binom{a+1}{a} s_{1,n} + (-1)^a n.\]

This recursive identity gives a formula for \(s_{a,n}\) in terms of \(s_{b,n}\) for \(b < a.\) It is the basis of many inductive arguments. In particular, the first pattern that one notices after deriving \(s_{a,n}\) for \(a=1,2,3\) is the leading terms \(\frac12 n^2, \frac13 n^3, \frac14 n^4.\) Here is an easy argument that the pattern continues:

For a positive integer \(a,\) \(s_{a,n}\) is a polynomial of degree \(a+1\) in \(n.\) Its leading term is \(\frac1{a+1} n^{a+1}.\)

Induction. The statement is true for \(a=1,\) and now suppose it is true for all positive integers less than \(a.\) Then solve the above recurrence for \(s_{a,n}\) to get

\[s_{a,n} = \frac1{a+1} n^{a+1} + c_{a-1} s_{a-1,n} + c_{a-2} s_{a-2,n} + \cdots + c_1 s_{1,n} + c_0 n,\]

where the \(c_i\) are some rational numbers.

Now by the inductive hypothesis, all of the terms except for the first term are polynomials of degree \(\le a\) in \(n,\) so the statement follows. \(_\square\)

Note the analogy to the continuous version of the sum: the integral \(\int_0^n x^a \, dx = \frac1{a+1}n^{a+1}.\) The lower-degree terms can be viewed as error terms in the approximation of the area under the curve \(y=x^a\) by the rectangles of width \(1\) and height \(k^a.\)

Faulhaber's Formula

Having established that \(s_{a,n} = \frac1{a+1} n^{a+1} +\text{(lower terms)},\) the obvious question is whether there is an explicit expression for the lower terms. It turns out that the terms can be expressed quite concisely in terms of the Bernoulli numbers, as follows:

Faulhaber's Formula:

\[\sum_{k=1}^n k^a = \frac1{a+1} \sum_{j=0}^{a} (-1)^j \binom{a+1}{j} B_j n^{a+1-j}.\]

That is, if \(i=a+1-j\) is a positive integer, the coefficient of \(n^i\) in the polynomial expression for the sum is \(\dfrac{(-1)^{a+1-i}}{a+1} \binom{a+1}{i} B_{a+1-i}.\)

Show that \(\sum\limits_{k=1}^n k^a = \frac1{a+1} n^{a+1} + \frac12 n^a + (\text{lower terms}).\)

This can be read off directly from Faulhaber's formula: the \(j=0\) term is \(\frac1{a+1}n^{a+1},\) and the \(j=1\) term is

\[\frac1{a+1} (-1)^1 \binom{a+1}1 B_1 n^a,\]

and since \(B_1 = -\frac12,\) this simplifies to \(\frac12 n^a.\) \(_\square\)

To compute \(\sum\limits_{k=1}^n k^4\) using Faulhaber's formula, write

\[\sum_{k=1}^n k^4 = \frac15 \sum_{j=0}^4 (-1)^j \binom{5}{j} B_j n^{5-j}\]

and use \(B_0 = 1, B_1 = -\frac12, B_2 = \frac16, B_3 = 0, B_4 = -\frac1{30}\) to get

\[\sum_{k=1}^n k^4 = \frac15 \left( n^5 + \frac52 n^4 + \frac{10}6 n^3 + 0 n^2 - \frac16 n\right) = \frac15 n^5 + \frac12 n^4 + \frac13 n^3 - \frac16 n.\]

This happens to factor as

\[\sum_{k=1}^n k^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}.\]

Note that the \((-1)^j\) sign only affects the term when \(j=1,\) because the odd Bernoulli numbers are zero except for \(B_1 = -\frac12.\)

The proof of the theorem is straightforward (and is omitted here); it can be done inductively via standard recurrences involving the Bernoulli numbers, or more elegantly via the generating function for the Bernoulli numbers.

See Also

  • Bernoulli Numbers

Cite as: Sum of n, n², or n³. Brilliant.org. Retrieved from https://brilliant.org/wiki/sum-of-n-n2-or-n3/

Sum of n, n², or n³ | Brilliant Math & Science Wiki (2024)

FAQs

Is the sum in math the answer? ›

When we add two or more numbers, the result or the answer we get can be defined as the SUM. The numbers that are added are called addends. In the above example, 6 and 4 are addends, and 10 is their sum. In other words, we can say the sum of 8 and 5 is 13 or 8 added to 5 is 13.

How to find the sum of n^2? ›

The sum of the first n squares, 1 + 4 + 9 + 16 + ... + n², is given by the formula ⅙n(n+1)(2n+1).

What does n mean in math? ›

The letter (N) is the symbol used to represent natural numbers. Natural numbers are also known as counting numbers, and they begin with the number 1 and continue to infinity (never ending), which is represented by three dots (...).

What is the rule of sum in math? ›

Rule of Sum - Statement:

If there are n n n choices for one action, and m m m choices for another action and the two actions cannot be done at the same time, then there are n + m n+m n+m ways to choose one of these actions.

What is the sum theory in math? ›

The sum of the interior angles in a triangle is supplementary. In other words, the sum of the measure of the interior angles of a triangle equals 180°. So, the formula of the triangle sum theorem can be written as, for a triangle ABC, we have ∠A + ∠B + ∠C = 180°.

What is sum called in math? ›

In mathematics, summation is the addition of a sequence of numbers, called addends or summands; the result is their sum or total.

What is the sum of n^3? ›

+ n 3 = ( n ( n +1)/2)2, which is the square of the n th triangle number. For example, 13+23+... +103=(10×11/2)2=552 = 3025.

What is n squared? ›

The usual notation for the square of a number n is not the product n × n, but the equivalent exponentiation n2, usually pronounced as "n squared". The name square number comes from the name of the shape. The unit of area is defined as the area of a unit square (1 × 1). Hence, a square with side length n has area n2.

How do you sum n numbers? ›

Sn = n(n+1)/2

Hence, this is the formula to calculate sum of 'n' natural numbers.

What does N mean in science? ›

newton, absolute unit of force in the International System of Units (SI units), abbreviated N. It is defined as that force necessary to provide a mass of one kilogram with an acceleration of one metre per second per second.

What does Z stand for in math? ›

The capital Latin letter Z is used in mathematics to represent the set of integers. Usually, the letter is presented with a "double-struck" typeface to indicate that it is the set of integers.

Is zero a real number? ›

Yes, 0 is a real number in math. By definition, the real numbers consist of all of the numbers that make up the real number line. The number 0 is at the center of the number line, so we know that 0 is a real number. Furthermore, 0 is a whole number, an integer, and a rational number.

What does ∑ mean? ›

The symbol ∑ indicates summation and is used as a shorthand notation for the sum of terms that follow a pattern.

What is the theory of summation? ›

Summation or sigma (∑) notation is a method used to write out a long sum in a concise way. This notation can be attached to any formula or function. For example, i=110 (i) is a sigma notation of the addition of finite sequence 1 + 2 + 3 + 4…… + 10 where the first element is 1 and the last element is 10.

What is the sum rule in biology? ›

The sum rule is applied when considering two mutually-exclusive outcomes that can result from more than one pathway. It states that the probability of the occurrence of one event or the other, of two mutually-exclusive events, is the sum of their individual probabilities.

Is the sum plus or minus? ›

The symbol we use for addition is + The answer to an addition problem is called the sum. The symbol we use for subtraction is – The answer to a subtraction problem is called the difference.

Is the sum the answer to a multiplication problem? ›

Answer and Explanation:

There is not a 'sum' in a multiplication problem. A 'sum' is the answer to an addition problem. In multiplication, the answer is called the 'product.

What is the sum of mean in math? ›

Mean = Sum of the Given Data/Total number of Data

To calculate the arithmetic mean of a set of data we must first add up (sum) all of the data values (x) and then divide the result by the number of values (n).

What is the answer to a sum called? ›

When two numbers are added together i.e. a+b, the result is called sum and the two numbers being added together are called addends. Similarly, when we subtract two number i.e. a-b, the result is called difference and a is called as subtrahend & b is called as minuend. Hope this helps!

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