Sums of squares, cubes, and fourth powers: formulas and proofs

The three formulas

For a positive integer $n$ ,

\boxed{\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}6,}

\boxed{\sum_{k=1}^n k^3=\frac{n^2(n+1)^2}4,}

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

The same formulas hold for $n=0$ if the sums are interpreted as empty sums. Here we derive them, rather than treating them as unrelated facts to memorize.

The common idea is to evaluate a sum of differences in two ways: telescope it, and expand each difference. We use the familiar identity $\sum_{k=1}^n k=n(n+1)/2$ .

Deriving the sum of squares

Start with

(k+1)^3-k^3=3k^2+3k+1.

Summing from $k=1$ to $n$ , the left side telescopes:

\sum_{k=1}^n\{(k+1)^3-k^3\}=(n+1)^3-1.

If $S_2=\sum_{k=1}^n k^2$ , the expanded right side is

3S_2+\frac32n(n+1)+n.

Equating the two evaluations and solving for $S_2$ gives

S_2=\frac{(n+1)^3-1-\frac32n(n+1)-n}{3}=\boxed{\frac{n(n+1)(2n+1)}6.}

The cancellation is explained in more detail in telescoping sums.

Deriving the sum of cubes

Use the next power:

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

Let $S_3=\sum_{k=1}^n k^3$ . Summation gives

(n+1)^4-1=4S_3+6S_2+4\frac{n(n+1)}2+n.

Substitute the square-sum formula:

(n+1)^4-1=4S_3+n(n+1)(2n+1)+2n(n+1)+n.

Rearranging and factoring yields

\boxed{S_3=\frac{n^2(n+1)^2}4=\left(\sum_{k=1}^n k\right)^2.}

Thus the sum of the first $n$ cubes is the square of the sum of the first $n$ positive integers.

Deriving the sum of fourth powers

Start with

(k+1)^5-k^5=5k^4+10k^3+10k^2+5k+1.

Writing $S_4=\sum_{k=1}^n k^4$ gives

(n+1)^5-1=5S_4+10S_3+10S_2+5\frac{n(n+1)}2+n.

Substitute the results already obtained:

\begin{aligned} (n+1)^5-1={}&5S_4+\frac52n^2(n+1)^2\\ &+\frac53n(n+1)(2n+1)+\frac52n(n+1)+n. \end{aligned}

Solving for $S_4$ and factoring the polynomial gives

\boxed{S_4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}.}

The pattern continues: the difference of consecutive $(r+1)$ st powers contains a leading term $(r+1)k^r$ and lower powers. Once their sums are known, the same procedure finds the sum of $r$ th powers.

Proving a known formula by induction

Derivation explains how to discover a formula. If the formula is already supplied or conjectured, mathematical induction can verify it efficiently.

For example, prove

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

At $n=1$ , both sides equal $1$ . Suppose the formula holds at a positive integer $m$ . Then

\begin{aligned} \sum_{k=1}^{m+1}k^3 &=\frac{m^2(m+1)^2}4+(m+1)^3\\ &=\frac{(m+1)^2}{4}\{m^2+4m+4\}\\ &=\frac{(m+1)^2(m+2)^2}{4}. \end{aligned}

This is the formula at $m+1$ , so the induction is complete. The base case and the use of the induction hypothesis are both explicit.

When expanding into power sums is unnecessary

Problem. Evaluate

\sum_{k=1}^n k(k+1)(k+2)(k+3).

Expanding gives a fourth-degree polynomial, so the formulas above could solve it. But the consecutive factors contain a shorter structure.

Let $P_k=k(k+1)(k+2)(k+3)$ . Multiplying by the next and previous factors gives

P_k(k+4)-P_k(k-1)=5P_k.

Therefore

P_k=\frac15\{k(k+1)(k+2)(k+3)(k+4)-(k-1)k(k+1)(k+2)(k+3)\}.

In the sum, the negative product for one index cancels the positive product for the previous index. Only the last positive term and the first negative term remain. The first negative term is zero because it contains $k-1=0$ at $k=1$ . Hence

\boxed{\sum_{k=1}^n k(k+1)(k+2)(k+3)=\frac{n(n+1)(n+2)(n+3)(n+4)}5.}

This calculation does not require memorizing the fourth-power formula. More generally, consecutive products often have a simple finite-difference antiderivative; look for that structure before expanding.

How to check a power-sum answer

Test a small value such as $n=1$ or $2$ . Then verify that the proposed expression $F(n)$ has $F(n)-F(n-1)=n^r$ and the correct starting value. These checks also explain why a correct formula holds for every integer $n$ , not just the first few.