Sums of powers of natural numbers

Beyond arithmetic progressions

You can already add the terms of an arithmetic or a geometric progression. Many series in JEE are neither. Look at $1^2 + 2^2 + 3^2 + \dots$ or $1 \times 3 + 2 \times 5 + 3 \times 7 + \dots$. The gaps between their terms keep changing, and so does the ratio.

Three formulas unlock most of these series. In sigma notation, $\sum_{k=1}^{n} k^2$ means $1^2 + 2^2 + \dots + n^2$.

The sum of the first n numbers

Write the sum forwards and backwards, one under the other:

Each column adds up to $n + 1$, and there are $n$ columns. So $2S = n(n + 1)$:

The sum of the squares

The trick here is a sum that collapses. Start from an identity that is true for every $k$:

Write it for $k = 1, 2, \dots, n$ and add all the lines.

1. The left side. $(2^3 - 1^3) + (3^3 - 2^3) + \dots + ((n + 1)^3 - n^3)$. Every middle cube appears once with a plus and once with a minus. Only $(n + 1)^3 - 1$ is left.
2. The right side. $3\sum k^2 + 3\sum k + n$.
3. Solve for the squares. $3\sum k^2 = (n + 1)^3 - 1 - \dfrac{3n(n + 1)}{2} - n$. Factorise and simplify.

Example 1. $1^2 + 2^2 + \dots + 10^2 = \dfrac{10 \times 11 \times 21}{6} = 385$.

The sum of the cubes

The same trick with $(k + 1)^4 - k^4$ gives the cubes. The answer is a surprise. It is the square of the first formula:

The picture shows why for $n = 4$. A square of side $1 + 2 + 3 + 4 = 10$ is cut into L-shaped bands. The bands have areas $1, 8, 27$ and $64$, which are the first four cubes.

Example 2. $1^3 + 2^3 + \dots + 10^3 = 55^2 = 3025$.

Notice that $\sum k^3 = \left(\sum k\right)^2$, but $\sum k^2$ is not $\left(\sum k\right)^2$. Do not mix them up.

Series with a polynomial kth term

Most JEE series of this kind need four steps.

1. Find the kth term. Write $T_k$ as a polynomial in $k$. Check it against the first two terms.
2. Split the sum. $\sum (ak^2 + bk + c) = a\sum k^2 + b\sum k + cn$.
3. Use the formulas. Put in the three sums.
4. Check. Put $n = 1$ and $n = 2$ into your answer.

Example 3. Find $1 \times 3 + 2 \times 5 + 3 \times 7 + \dots$ to $n$ terms.

1. The kth term. The first factors are $1, 2, 3, \dots$ and the second are $3, 5, 7, \dots$. So $T_k = k(2k + 1) = 2k^2 + k$.
2. Split and sum. $2 \times \dfrac{n(n + 1)(2n + 1)}{6} + \dfrac{n(n + 1)}{2} = \dfrac{n(n + 1)(4n + 5)}{6}$.
3. Check. $n = 1$ gives $\dfrac{1 \times 2 \times 9}{6} = 3$, and $n = 2$ gives $\dfrac{2 \times 3 \times 13}{6} = 13 = 3 + 10$.

Example 4 (a sum that does not start at 1). Find $6^3 + 7^3 + \dots + 15^3$. Take the cubes up to 15, then remove the cubes up to 5:

Practise with new numbers

Try it in Python

The program adds the powers one at a time and prints each formula beside the sum.

Common mistakes

• Squaring the wrong sum. $\sum k^3 = \left(\sum k\right)^2$, but $\sum k^2$ is a different formula.
• A wrong kth term. Always test $T_k$ against the first two terms of the series.
• A constant term. $\sum_{k=1}^{n} c = cn$, not $c$.
• A sum that does not start at 1. Take the sum up to the last term, then subtract the part you do not want.

Remember these

• $\sum k = \dfrac{n(n + 1)}{2}$, $\sum k^2 = \dfrac{n(n + 1)(2n + 1)}{6}$ and $\sum k^3 = \left[\dfrac{n(n + 1)}{2}\right]^2$.
• Expand the kth term, split the sum, then use the three formulas.
• Check every answer with $n = 1$ and $n = 2$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find $1^2 + 2^2 + \dots + 20^2$.

2. Find $1^3 + 2^3 + \dots + 8^3$.

3. Write $\sum_{k=1}^{n} (3k - 1)^2$ as a polynomial in $n$.

4. Show that $1^3 + 2^3 + \dots + n^3$ is always a perfect square.