2025-06-17-mathsproofs

Proof of the Formula for the Sum of an Arithmetic Series

Definition

An arithmetic series is the sum of terms in an arithmetic sequence, where each term differs from the previous by a constant value called the common difference (d).

The general form is: $S_n=a+(a+d)+(a+2d)+(a+3d)+\cdots +(a+(n-1)d)$

Where:

• $a$ = first term
• $d$ = common difference
• $n$ = number of terms
• $S_n$ = sum of the first $n$ terms

The Proof

Step 1: Write out the sum in standard order $S_n=a+(a+d)+(a+2d)+(a+3d)+\cdots +(a+(n-1)d)$

Step 2: Write out the same sum in reverse order $S_n=(a+(n-1)d)+(a+(n-2)d)+(a+(n-3)d)+\cdots +(a+d)+a$

Step 3: Add the two equations term by term $S_n+S_n=[a+(a+(n-1)d)]+[(a+d)+(a+(n-2)d)]+[(a+2d)+(a+(n-3)d)]+\cdots$

Step 4: Simplify each bracketed pair Notice that each pair sums to the same value:

• First pair: $a+(a+(n-1)d)=2a+(n-1)d$
• Second pair: $(a+d)+(a+(n-2)d)=2a+d+(n-2)d=2a+(n-1)d$
• Third pair: $(a+2d)+(a+(n-3)d)=2a+2d+(n-3)d=2a+(n-1)d$

Every pair equals $2a+(n-1)d$, and we have $n$ such pairs.

Step 5: Solve for $S_n$ $2S_n=n\cdot [2a+(n-1)d]$

$S_n=\frac{n}{2}\cdot [2a+(n-1)d]$

Alternative Forms

The formula can also be written as:

Form 1: $S_n=\frac{n}{2}[2a+(n-1)d]$

Form 2: $S_n=\frac{n}{2}(a+l)$ where $l$ is the last term

Form 3: $S_n=na+\frac{n(n-1)d}{2}$

Example

Find the sum of the first 10 terms of the arithmetic series: 3, 7, 11, 15, …

• $a=3$ (first term)
• $d=4$ (common difference)
• $n=10$ (number of terms)

$S_{10}=\frac{10}{2}[2(3)+(10-1)(4)]=5[6+36]=5\times 42=210$

Proof of the Formula for the Sum of a Geometric Series

Let’s now prove the formula. We want to find the sum

$S_N=a_1+a_2+a_3+\cdots +a_N.$

The formula for the $n$th term is

$a_n=a_1\cdot r^{n-1}.$

Plugging this expression into $S_N$, we get

$S_N=a_1+a_1\cdot r+a_1\cdot r^2+\cdots +a_1\cdot r^{N-1}.$

Now, let’s write down $-r{S}_N$ underneath the expression for $S_N$:

\begin{align} S_N &= a_1 + a_1 \cdot r + a_1 \cdot r^2 + a_1 \cdot r^3 + \cdots + a_1 \cdot r^{N-1}\\ -r \cdot S_N &= \phantom{a_1} - a_1 \cdot r - a_1 \cdot r^2 - a_1 \cdot r^3 - \cdots - a_1 \cdot r^{N-1} - a_1 \cdot r^N. \end{align}

Summing the two equations above, we get

$S_N-r\cdot S_N=a_1+(a_1\cdot r-a_1\cdot r)+(a_1\cdot r^2-a_1\cdot r^2)+\cdots +(a_1\cdot r^{N-1}-a_1\cdot r^{N-1})-a_1\cdot r^N.$

We see that all terms of $S_N$ are canceled but the first and the last. So we get

$S_N-r\cdot S_N=a_1-a_1\cdot r^N$

$S_N(1-r)=a_1(1-r^N)$

$S_N=a_1\frac{(1-r^N)}{1-r}.$

Sum to infinity

$S_{\infty }=\frac{a_1}{1-r}\text{for}|r|<1$ This comes from the above formula for Sum to a number. We use a limit to determine the following. As $|r|<1$ then $r^n\to 0$ as $n\to \infty$. Therefore, as $n\to \infty$ we have

$S_n\to a_1\cdot \frac{1-0}{1-r}=a_1\cdot \frac{1}{1-r}=\frac{a_1}{1-r}=S_{\infty }$

Sum of Sine and Cosine

We want to prove the following identities:

• $\cos (x+y)=\cos x\cos y-\sin x\sin y$
• $\sin (x+y)=\sin x\cos y+\cos x\sin y$

We start with Euler’s formula, which states:

$$e^{i\theta }=\cos \theta +i\sin \theta$$

With Euler”s we can write a complex number $z$ in 3 equivalent ways, letting $r=|z|$ and $\theta =arg(z)$

	Cartesian	Polar	Exponential
$z$	$x+iy$	$r(cos\theta +isin\theta )$	$r{e}^{i\theta }$

Let us apply this formula to the sum $x+y$

$$e^{i(x+y)}=e^{ix}\cdot e^{iy}$$

Now, using Euler’s formula on both sides:

$$\cos (x+y)+i\sin (x+y)=(\cos x+i\sin x)(\cos y+i\sin y)$$

Expand the right-hand side:

$$=\cos x\cos y+i\cos x\sin y+i\sin x\cos y+i^2\sin x\sin y$$ $$=\cos x\cos y+i(\cos x\sin y+\sin x\cos y)-\sin x\sin y$$ $$=(\cos x\cos y-\sin x\sin y)+i(\cos x\sin y+\sin x\cos y)$$

Now equate both sides:

$$\cos (x+y)+i\sin (x+y)=(\cos x\cos y-\sin x\sin y)+i(\cos x\sin y+\sin x\cos y)$$

Now equating the real parts and imaginary parts:

• Real parts:

  $$\cos (x+y)=\cos x\cos y-\sin x\sin y$$

• Imaginary parts:

  $$\sin (x+y)=\cos x\sin y+\sin x\cos y$$

Thus, the sum identities for sine and cosine are proved using Euler’s formula.