# Math Notes about Sum Formulas Dhemy

July 9, 2022 - 5 min read

I’m working on enhancing my skills, expanding my knowledge and increasing my chances to pass technical interviews. Mathematics plays an important role in problem-solving. Hence, this notes! This blog post is about Sum Formulas.

Each sum of the form: $$\sum_{x=1}^{n} x^{k} = 1^{k} + 2^{k} + 3^{k} + .. + n^{k}$$ where k is a positive integer, has a closed-form formula that is a polynomial of degree (k + 1).

Triangular number

$\sum_{x=1}^{n} x = 1 + 2 + 3 + ... + n = \frac{n (n + 1)}{2}$

Square pyramidal number

$\sum_{x=1}^{n} x^{2} = 1^{2} + 2^{2} + 3^{2} + ... + n^{2} = \frac{n (n + 1) (2n + 1)}{6}$

There is a general formula for such sums, called Faulhaber’s Formula.

In the next lines, I’ll use mathematical induction to proof the Triangular number and the Square pyramidal number formulas.

## Triangular number.

$$\sum_{x=1}^{n} x = 1 + 2 + 3 + ... + n = \frac{n (n + 1)}{2}$$

Base step: Base case, for n = 1

$L.H.S = 1$ $R.H.S = \frac{n (n + 1)}{2} = \frac{1 ( 1 + 1)}{2} = \frac{2}{2} = 1$ $L.H.S = R.H.S = 1$

Inductive hypothesis Step The equation is true for an arbitrary integer k

$1 + 2 + 3 + ... + k = \frac{k(k+1)}{2}$

Induction Step

Based on the inductive hypothesis, let’s proof the equation if true for n = k + 1

$L.H.S = 1 + 2 + 3 ... + k + k + 1$ $= \frac{k(k+1)}{2} + (k + 1)$ $= \frac{k(k+1)}{2} + \frac{2(k + 1)}{2}$ $= \frac{k^2 + k + 2k + 2}{2}$ $= \frac{k^2 + 3k + 2}{2}$ $= \frac{(k+1)(k+2)}{2}$ $= \frac{(k + 1)(k+1+1)}{2}$

For n = k + 1

$R.H.S = \frac{n(n + 1)}{2} = \frac{(k+1)(K + 1 + 1)}{2}$ $L.H.S = R.H.S \#$

## Square pyramidal number

$$\sum_{x=1}^{n} x^{2} = 1^{2} + 2^{2} + 3^{2} + ... + n^{2} = \frac{n (n + 1) (2n + 1)}{6}$$

Base step: Base case, for n = 1

$L.H.S = 1^2 = 1$ $R.H.S = \frac{1(1+1)(2 \times 1 + 1)}{6} = \frac{2 \times 3}{6} = 1$ $L.H.S = R.H.S$

Inductive hypothesis Step The equation is true for an arbitrary integer k

$1^2 + 2^2 + 3^2 + ... + k^2 = \frac{k(k+1)(2k+1)}{6}$

Induction Step

Based on the inductive hypothesis, let’s proof the equation if true for n = k + 1

$L.H.S = 1^2 + 2^2 + 3^2 + ... + k^2 + (k+1)^2$ $= \frac{k(k+1)(2k+1)}{6} + (k+1)^2$ $= \frac{k(k+1)(2k+1)}{6} + \frac{6(k+1)^2}{6}$

Given the R.H.S:

$R.H.S = \frac{n(n+1)(2n+1)}{6} = \frac{(k+1)(k+2)(2k+3)}{6}$

So, we can ignore the 6 in the denominator from both sides for now:

${L.H.S}\times{6} = k(k+1)(2k+1) + 6(k+1)^2$ $= (k^2 + k)(2k+1) + 6(k^2+2k+1)$ $= 2k^3 + k^2 + 2k^2 + k + 6k^2 + 12k + 6$ ${L.H.S}\times{6} = 2k^3 + 9k^2 + 13k + 6$
${R.H.S}\times{6} = (k+1)(k+2)(2k+3)$ $= (k^2 + 3k + 2)(2k+3)$ $= 2k^3 + 6k^2 + 3k^2 + 4k + 9k + 6$ ${R.H.S}\times{6} = 2k^3 + 9k^2 + 13k + 6$
$L.H.S = \frac{2k^3 + 9k^2 + 13k + 6}{6}$ $R.H.S = \frac{2k^3 + 9k^2 + 13k + 6}{6}$ $R.H.S = L.H.S \#$

## Number Progressions

A progression, which is also known as a sequence, is nothing but a pattern of numbers. For example, 3, 6, 9, 12 is a progression because there is a pattern observed where every number here is obtained by adding 3 to its previous number.

There are four types of the most common progressions.

### Arithmetic Progression

Arithmetic Progression (AP) is a sequence of numbers where the difference between any two consecutive numbers is constant.

For Example: 3, 7, 11, 15, is an AP where the common difference is 4.

For an AP:

• a is the first term.
• l is the last term.
• d is the common difference.
• The n term could be found as follows: $$a_{n} = a + (n - 1) d$$
• We can find the sum as follows: $$S = \frac{n(a + l)}{2}$$

Let’s proof this formula:

The sum of an arithmetic is $$S = \frac{n(a + l)}{2}$$

Suppose $$a_{1}, a_{2}, a_{3}, ..., a_{n}$$ be an arithmetic progression whose first term is a and the common difference is d.

Then:

$a_{1} = a$ $a_{2} = a + d$ $a_{3} = a + 2d$ $a_{n} = a + (n - 1) d$
$S_{n} = a + (a + d) + (a + 2d) + ... + \{a + (n - 3)d\} + \{a + (n - 2)d\} + \{a + (n - 1)d\} \;\;\;\;\;\;\;\; (i)$

And with a simple trick of reversing the order:

$S_{n} = \{a + (n - 1)d\} + \{a + (n - 2)d\} + \{a + (n - 3)d\} + ... + (a + 2d) + (a + d) + a \;\;\;\;\;\;\;\; (ii)$

by adding $$(i)$$ and $$(ii)$$

Term (i) Term (ii) Sum
a $$a + (n - 1)d$$ $$2a + (n-1)d$$
a + d $$a + (n - 2)d$$ $$2a + (n-1)d$$
a + 2d $$a + (n - 3)d$$ $$2a + (n-1)d$$
$$2S_{n} = n\{2a + (n-1)d\}$$
$2S_{n} = \{2a + (n-1)d\} + \{2a + (n-1)d\} + ... + \{2a + (n-1)d\}$
$2S_{n} = n\{2a + (n-1)d\}$ $2S_{n} = n\{a + a + (n-1)d\}$ $2S_{n} = n(a + l)$ $S_{n} = \frac{n(a+l)}{2} \#$

### Geometric Progression

Geometric Progression (GP) is a sequence of numbers where the ratio between any two consecutive numbers is constant.

$S = \frac{l\times k - a}{k - 1}$

### Harmonic Progression

A series of numbers is said to be in harmonic sequence if the reciprocals (the inverse) of all the elements of the sequence form an arithmetic sequence.

For example: $$\sum_{x=1}^{n}{\frac{1}{x}} = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}$$

### Fibonacci Progression

Fibonacci numbers form an interesting sequence of numbers in which each element is obtained by adding two preceding elements and the sequence starts with 0 and 1.

Sequence is defined as, $$F_{0} = 0$$ and $$F_{1} = 1$$ and $$F_{n} = F_{n-1} + F_{n-2}$$