Sequences and Series • Topic 3 of 3

Special Series

Beyond APs and GPs, three particular sums appear so often that their closed formulas are worth memorising. They are the sums of the first $n$ natural numbers, their squares, and their cubes. A special series is one whose terms follow a simple power pattern, letting us replace a long addition with a single tidy expression.

Sum of the first $n$ natural numbers — this is just an AP with $a = 1$, $d = 1$, so Gauss's pairing gives:

$$\sum_{k=1}^{n} k = 1 + 2 + 3 + \dots + n = \dfrac{n(n+1)}{2}$$

Sum of the first $n$ squares:

$$\sum_{k=1}^{n} k^2 = 1^2 + 2^2 + \dots + n^2 = \dfrac{n(n+1)(2n+1)}{6}$$

Sum of the first $n$ cubes — strikingly, this is exactly the square of the first formula:

$$\sum_{k=1}^{n} k^3 = 1^3 + 2^3 + \dots + n^3 = \left[\dfrac{n(n+1)}{2}\right]^2 = \left(\sum_{k=1}^{n} k\right)^2$$

These three are the toolkit; their real power shows when you combine them. Because summation is linear, you can split and scale any polynomial term and apply each formula separately:

$$\sum_{k=1}^{n}\left(\alpha k^2 + \beta k + \gamma\right) = \alpha\sum k^2 + \beta\sum k + \gamma\, n$$

Here is the full reference set side by side:

SeriesClosed form$n = 4$ check
$\sum k$$\dfrac{n(n+1)}{2}$$1+2+3+4 = 10$
$\sum k^2$$\dfrac{n(n+1)(2n+1)}{6}$$1+4+9+16 = 30$
$\sum k^3$$\left[\dfrac{n(n+1)}{2}\right]^2$$1+8+27+64 = 100$

The way to use these on an unfamiliar series is the nth-term method: find a formula $a_k$ for the general term, then sum it. For instance, the series $1\cdot 2 + 2\cdot 3 + 3\cdot 4 + \dots$ has $a_k = k(k+1) = k^2 + k$, so its sum is $\sum k^2 + \sum k$, which you can evaluate immediately from the table.

Deeper Insight — find the term, then the formulas do the rest: The single most transferable skill in this topic is learning to read a series, spot the rule generating its general term $a_k$, and rewrite that term as a combination of powers of $k$. Once a sum is expressed as $\sum(\alpha k^2 + \beta k + \gamma)$, the three closed forms turn an apparently hard problem into routine arithmetic — no clever trick required, just bookkeeping. The cube identity $\sum k^3 = (\sum k)^2$ deserves a moment of wonder: it says the sum of the first $n$ cubes is a perfect square, and not just any square but the square of the triangular number $\tfrac{n(n+1)}{2}$. This is the kind of unexpected harmony that makes the subject worth studying, and it has a beautiful geometric proof involving nested square frames. Mastering the nth-term method here pays off far beyond this chapter: the same idea underlies the method of differences, telescoping sums, and ultimately the definite integral, where summing infinitely many tiny terms is exactly what you will be doing in Class 12.

Sum of the first n natural numbers as a triangular staircase of unit blocks Σ k = n(n+1) / 2 (triangular staircase, n = 5) 12345 = 15 The three special series formulas grouped together Three formulas to memorise Σ kn(n+1) / 2 Σ k²n(n+1)(2n+1) / 6 Σ k³[ n(n+1) / 2 ]²
1
Worked Example
Find the sum of the first $50$ natural numbers.
Solution
  1. Use $\sum_{k=1}^{n} k = \dfrac{n(n+1)}{2}$ with $n = 50$.
  2. $= \dfrac{50 \times 51}{2} = \dfrac{2550}{2}$.
  3. $= 1275$.

Answer: $1275$.

2
Worked Example
Find $\sum_{k=1}^{10} k^2$, the sum of the squares of the first $10$ natural numbers.
Solution
  1. Use $\sum_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6}$ with $n = 10$.
  2. $= \dfrac{10 \times 11 \times 21}{6} = \dfrac{2310}{6}$.
  3. $= 385$.

Answer: $385$.

3
Worked Example
Find the sum of the cubes of the first $6$ natural numbers.
Solution
  1. Use $\sum_{k=1}^{n} k^3 = \left[\dfrac{n(n+1)}{2}\right]^2$ with $n = 6$.
  2. $\dfrac{6 \times 7}{2} = 21$, so the sum is $21^2$.
  3. $= 441$.

Answer: $441$.

4
Worked Example
Find the sum of the series $1\cdot 2 + 2\cdot 3 + 3\cdot 4 + \dots$ up to $n$ terms.
Solution
  1. The general term is $a_k = k(k+1) = k^2 + k$.
  2. $\sum_{k=1}^{n} a_k = \sum k^2 + \sum k = \dfrac{n(n+1)(2n+1)}{6} + \dfrac{n(n+1)}{2}$.
  3. Take out $\dfrac{n(n+1)}{6}$: $= \dfrac{n(n+1)}{6}\left[(2n+1) + 3\right] = \dfrac{n(n+1)(2n+4)}{6}$.
  4. $= \dfrac{n(n+1)(n+2)}{3}$.

Answer: $\dfrac{n(n+1)(n+2)}{3}$.

5
Worked Example
Evaluate $\sum_{k=1}^{20} (3k^2 - 2k + 1)$.
Solution
  1. Split using linearity: $3\sum k^2 - 2\sum k + \sum 1$ for $n = 20$.
  2. $\sum k^2 = \dfrac{20 \times 21 \times 41}{6} = 2870$; $\sum k = \dfrac{20 \times 21}{2} = 210$; $\sum_{k=1}^{20} 1 = 20$.
  3. $3(2870) - 2(210) + 20 = 8610 - 420 + 20$.
  4. $= 8210$.

Answer: $8210$.

6
Worked Example
Find the sum of the series $2^2 + 4^2 + 6^2 + \dots + (2n)^2$.
Solution
  1. The general term is $(2k)^2 = 4k^2$ for $k = 1$ to $n$.
  2. $\sum_{k=1}^{n} 4k^2 = 4\sum k^2 = 4 \times \dfrac{n(n+1)(2n+1)}{6}$.
  3. $= \dfrac{2n(n+1)(2n+1)}{3}$.

Answer: $\dfrac{2n(n+1)(2n+1)}{3}$.

Key Points

  • $\sum_{k=1}^{n} k = \dfrac{n(n+1)}{2}$ — the first $n$ natural numbers.
  • $\sum_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6}$ — the first $n$ squares.
  • $\sum_{k=1}^{n} k^3 = \left[\dfrac{n(n+1)}{2}\right]^2$ — the first $n$ cubes, equal to $\left(\sum k\right)^2$.
  • Summation is linear: split $\sum(\alpha k^2 + \beta k + \gamma)$ and apply each formula separately.
  • For an unfamiliar series, find the nth term $a_k$ first, rewrite it in powers of $k$, then sum.
Tap an option to check your answer0 / 4
Q1.$\displaystyle\sum_{k=1}^{n} k=$
Explanation: Sum of first $n$ natural numbers.
Q2.$\displaystyle\sum_{k=1}^{n} k^2=$
Explanation: Sum of squares.
Q3.$\displaystyle\sum_{k=1}^{n} k^3=$
Explanation: Sum of cubes is the square of the sum.
Q4.$1+2+\dots+10=$
Explanation: $\tfrac{10\cdot11}{2}=55$.