Something Interesting in Arithmetic Sequences

$1+2+3+\cdots +n$

I believe you are familiar with this formula. The German mathematician Gauss is said to have studied the formula at the age of three. To this day, there are many explanations. The most widely used method, which Gauss used in those days, was to add the first and last terms, the second and the penultimate.

For the arithmetic progression $1,2,3,\cdots ,n$, there are $\displaystyle \sum_{i=1}^n i=\dfrac{n^2}{2}+\dfrac{n}{2}$ For the arithmetic progression $a_1,a_2,a_3\cdots a_n$, if $a_n-a_{n-1}=d$, there are $a_n=a_1+(n-1)d$ $d=\dfrac{a_m-a_n}{m-n}$ $a_1+a_n=a_2+a_{n-1}=\cdots =a_k+a_{n-k+1}$ $\displaystyle \sum_{i=1}^n a_i=\dfrac{(a_1+a_n)\times n}{2}=a_1\times n+\dfrac{n(n-1)\times d}{2}$

For example:

$(1)$ Find $1+2+3+\cdots +50$. $(2)$ Find $1+2+3+\cdots +100$

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$\begin{aligned} (1) S &=1+2+3+\cdots +50 \\ &=(1+50)+(2+49)+(3+48)+\cdots \\ &=51\times \dfrac{50}{2} \\ &=1275 \end{aligned}$

$\begin{aligned} (2) S &=1+2+3+\cdots +100 \\ &=(1+100)+(2+99)+(3+98)+\cdots \\ &=101\times \dfrac{100}{2} \\ &=5050 \end{aligned}$

Find $1+2+3+\cdots +101$

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$\begin{aligned} S &=1+2+3+\cdots +101 \\ &=\dfrac{1}{2}\times [(1+2+3+\cdots +101)+(101+100+99+\cdots +1)] \\ &=\dfrac{1}{2}\times [(1+101)+(2+100)+(3+99)+\cdots] \\ &=\dfrac{1}{2}\times (1+101)\times 101 \\ &=5151 \end{aligned}$

The following picture also proves our derivation. This image comes from Basic Mathematics. In the picture, $S=1+2+3+\cdots +n=\dfrac{n^2}{2}+\dfrac{n}{2}$. We can see that the equation of line in the figure is $y=x$. It also shows that there is a close relationship between arithmetic progression and the Linear function $y=kx+b$.

$a_n=a_1+(n-1)d$ and $y=kx+b$

When the teacher talks about this, someone always asks this question. In fact, this problem can be solved by the general formula of arithmetic progression. For any arithmetic sequence $a_n$,there are

$a_1+a_n=a_2+a_{n-1}=\cdots =a_k+a_{n-k+1}$

The sum of arithmetic sequence

The above example is an arithmetic sequence with a tolerance equal to one. For arithmetic progression

$a_1,a_2,a_3\cdots a_n$

We can use the above idea to solve the problem. First, we tried to use algebraic methods.

$\displaystyle \sum_{i=1}^n a_i=\dfrac{1}{2}[(a_1+a_n)+(a_2+a_{n-1})+\cdots]=\dfrac{(a_1+a_n)\times n}{2}$

If we substitute $a_n$, we can get:

$\displaystyle \sum_{i=1}^n a_i=\dfrac{[2a_1+(n-1)d]\times n}{2}=a_1\times n+\dfrac{n(n-1)\times d}{2}$

There is another way to do this. Thanks to my friend Alex Dixon here, I have modified his method slightly, as follows:

$\begin{aligned} \because a_n &=a_1+(n-1)d \\ \therefore \displaystyle \sum_{i=1}^n a_i &=[a_1+(1-1)d]+[a_1+(2-1)d]+[a_1+(3-1)d]+\cdots +[a_1+(n-1)d] \\ &=a_1\times n+[d+2d+3d+\cdots +(n-1)d] \\ &=a_1\times n+[1+2+3+\cdots +(n-1)]d \\ &=a_1\times n+\dfrac{n(n-1)\times d}{2} \end{aligned}$

And once again, we can prove it with the pattern above. But it needs to be changed a little bit. I'm sorry that I can't submit pictures here. Those who are interested can draw a picture by yourselves. And you'll see that it becomes a complete trapezoid. And using mathematical induction, of course. Like this:

When $n=1$, $S_1=a_1$, $S_2=a_1\times 1+\dfrac{1\times (1-1)\times d}{2}=a_1$. There's $S_1=S_2$, so it works.

If $n=k$, there's $\displaystyle \sum_{i=1}^k a_i=a_1\times k+\dfrac{kd(k-1)}{2}$.

When $n=k+1$, we can get

$\begin{aligned} \displaystyle \sum_{i=1}^{k+1} a_i &=a_1\times k+\dfrac{k(k-1)\times d}{2}+a_1+kd \\ &=a_1\times (k+1)+\dfrac{k^2d-kd}{2}+kd \\ &=a_1\times (k+1)+\dfrac{k^2d+kd}{2} \\ &=a_1\times (k+1)+\dfrac{(k+1)[(k+1)-1]\times d}{2} \end{aligned}$