Something Interesting in Arithmetic Sequences

1+2+3++n1+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,,n1,2,3,\cdots ,n, there are i=1ni=n22+n2\displaystyle \sum_{i=1}^n i=\dfrac{n^2}{2}+\dfrac{n}{2} For the arithmetic progression a1,a2,a3ana_1,a_2,a_3\cdots a_n, if anan1=da_n-a_{n-1}=d, there are an=a1+(n1)da_n=a_1+(n-1)d d=amanmnd=\dfrac{a_m-a_n}{m-n} a1+an=a2+an1==ak+ank+1a_1+a_n=a_2+a_{n-1}=\cdots =a_k+a_{n-k+1} i=1nai=(a1+an)×n2=a1×n+n(n1)×d2\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)(1) Find 1+2+3++501+2+3+\cdots +50. (2)(2) Find 1+2+3++1001+2+3+\cdots +100

(1)S=1+2+3++50=(1+50)+(2+49)+(3+48)+=51×502=1275\begin{aligned} (1) S &=1+2+3+\cdots +50 \\ &=(1+50)+(2+49)+(3+48)+\cdots \\ &=51\times \dfrac{50}{2} \\ &=1275 \end{aligned}

(2)S=1+2+3++100=(1+100)+(2+99)+(3+98)+=101×1002=5050\begin{aligned} (2) S &=1+2+3+\cdots +100 \\ &=(1+100)+(2+99)+(3+98)+\cdots \\ &=101\times \dfrac{100}{2} \\ &=5050 \end{aligned}

From the above example, we can get 1+2+3++n=n22+n21+2+3+\cdots +n=\dfrac{n^2}{2}+\dfrac{n}{2}. But this explanation is flawed. When the number of terms is an odd number, there is always a number in the middle that cannot be paired with other Numbers. If I had to figure out the middle number, it would have been too much trouble. Like this example. We can do it the way we did it above, and there are many ways to do it. But let's think about it another way. We can eliminate this trouble by constructing two sequences 1,2,32n+1{1,2,3\cdots 2n+1} and 2n+1,2n,2n11{2n+1,2n,2n-1\cdots 1}, and adding the nthn^{th} term of the first sequence to the nthn^{th} term of the second sequence.

Find 1+2+3++1011+2+3+\cdots +101

S=1+2+3++101=12×[(1+2+3++101)+(101+100+99++1)]=12×[(1+101)+(2+100)+(3+99)+]=12×(1+101)×101=5151\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++n=n22+n2S=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=xy=x. It also shows that there is a close relationship between arithmetic progression and the Linear function y=kx+by=kx+b.

an=a1+(n1)da_n=a_1+(n-1)d and y=kx+by=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 ana_n,there are

a1+an=a2+an1==ak+ank+1a_1+a_n=a_2+a_{n-1}=\cdots =a_k+a_{n-k+1}
We combined Linear Function with arithmetic sequence, and set a1=k+ba_1=k+b, an=kn+ba_n=kn+b. So we can also understand 0n(kx+b)dx+12kn=i=1nai\displaystyle\int_0^n (kx+b) \mathrm{d}x+\dfrac{1}{2}kn=\displaystyle\sum_{i=1}^n a_i. This needs to do a lot of tedious promotion, here I will not write. 0n(kx+b)dx+12kn=(kx+b)dx0n+12kn=12kx2+bx+C0n+12kn=12kn2+bn+12kn=n2(kn+2b+k)=n2[(kn+b)+(k+b)]=(a1+an)n2\begin{aligned} \int_0^n (kx+b) \mathrm{d}x+\dfrac{1}{2}kn &=\left.\int (kx+b) \mathrm{d}x\right|_0^n+\dfrac{1}{2}kn \\ &=\left.\dfrac{1}{2}kx^2+bx+\mathrm {C}\right|_0^n+\dfrac{1}{2}kn \\ &=\dfrac{1}{2}kn^2+bn+\dfrac{1}{2}kn \\ &=\dfrac{n}{2}(kn+2b+k) \\ &=\dfrac{n}{2}[(kn+b)+(k+b)] \\ &=\dfrac{(a_1+a_n)n}{2} \end{aligned}

The sum of arithmetic sequence

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

a1,a2,a3ana_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.

i=1nai=12[(a1+an)+(a2+an1)+]=(a1+an)×n2\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 ana_n, we can get:

i=1nai=[2a1+(n1)d]×n2=a1×n+n(n1)×d2\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:

an=a1+(n1)di=1nai=[a1+(11)d]+[a1+(21)d]+[a1+(31)d]++[a1+(n1)d]=a1×n+[d+2d+3d++(n1)d]=a1×n+[1+2+3++(n1)]d=a1×n+n(n1)×d2\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=1n=1, S1=a1S_1=a_1, S2=a1×1+1×(11)×d2=a1S_2=a_1\times 1+\dfrac{1\times (1-1)\times d}{2}=a_1. There's S1=S2S_1=S_2, so it works.

If n=kn=k, there's i=1kai=a1×k+kd(k1)2\displaystyle \sum_{i=1}^k a_i=a_1\times k+\dfrac{kd(k-1)}{2}.

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

i=1k+1ai=a1×k+k(k1)×d2+a1+kd=a1×(k+1)+k2dkd2+kd=a1×(k+1)+k2d+kd2=a1×(k+1)+(k+1)[(k+1)1]×d2\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}

Note by Edward Christian
1 year, 10 months ago

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For an arithmetic sequence, an+b the equation can be written as x1 + x2 + x3 +…+ xn = (a+b) + (2a + b) + (3a + b) +…+ (an+b) => (a + 2a + 3a +…+ an) + (b + b + b +… to n terms) = a[1 + 2 + 3 +…+n] + nb = [a(n(n+1))/2] + nb = [an^2+ an]/2+ nb = n/2 [an+a+2b] = n/2 [an+a+b+b] = n/2 [(a+b)+(an+b)] In this sequence, (a+b) = x1 , and (an+b) = xn So, this equation can be written in terms of first and last term as, x1 + x2 + x3 +…+ xn = n/2 [x1+ xn]

Alex Dixon - 1 year, 9 months ago

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Emm...I think this is a great idea.CanI add it to my note?

Edward Christian - 1 year, 9 months ago

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We can derive more equations from the above. This equation can rewritten using the first and last terms of the sequence. You know, we can use different forms of the same equation in different problems. That's the interesting fact about mathematics. It's interesting, isn't it?

Alex Dixon - 1 year, 10 months ago

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Yes, Sir. The above problem is actually a special arithmetic sequence. And if we keep going, we're going to get a lot more interesting results. But I have to admit that, first of all, I wrote this NOTE in a hurry. Second, this NOTE is mainly intended to state an idea, not a result. Thirdly, I am only a middle school student with limited math skills. These are the only things I can say in a rigorous and methodical way. Although I have a lot of ideas, I am not very mature enough to write them here. If you have a better idea, please let me know.

Edward Christian - 1 year, 9 months ago

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Of course, you can

Alex Dixon - 1 year, 9 months ago

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You're welcome Edward

Alex Dixon - 1 year, 9 months ago

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How can you be such a great mathematician you’re only 13!

Joshua Olayanju - 1 year ago

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I am familiar with the theorem but have never viewed it like this.

Joshua Olayanju - 1 year ago

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