Quickly derive the sum formula of geometric sequence
One of the charms of n-ary numbers-inspired by the principles of computer composition
Preface
I still can’t remember the summation formula of a geometric sequence. Maybe this is also the pain point of most people. I think mathematics is like Chinese or English. Once more professional or novel idioms are used, the brain It's a short-circuit. For those who want to overcome such difficulties, I think this blog may be of some help to you. This is the first blog of the blogger. However, his level is limited. If there is any problem, I still hope Haihan...
The general term sequence of the summation sequence expansion form of the geometric sequence is as a n = 2 n
Then the first n terms and formula of the geometric sequence of this general term can be expressed as S n = 2 1 +2 2 +2 3 +2 4 +……+2 n, right?
Therefore, you only need to know this formula to expand it like this.
For example of n-base number, suppose the first 4 items and
It is 1111 2 =2 1 +2 2 +2 3 +2 4 to express in binary numbers . Seeing this, whether everyone has a glimpse, and listen to me...
The above binary number, if +1, becomes 10000 2 =2 5 , then the result of the above geometric sequence is 2 5 -1=32-1=31, if the sum of the above 4 items is changed to n Term sum, then the final geometric sequence result, whether it also becomes S n =2 n+1 -1 This simply derives the first n term sum of the geometric sequence with a n = 2 n as the general term
Examples of other n-base numbers In order to make everyone more clear, this time the general term is set as a n = 4 n
So how do you do it now? Let’s use a quaternary number to represent it. Let’s suppose that the sum of the first four terms is also the first, that is, 3333 4 = 3*(4 1 +4 2 +4 3 +4 4 ). At this point, I believe everyone wants to do it by themselves. I gave it a try, instead of continuing to look at it with great interest. I also think that everyone is very smart and will be able to deduce it soon. Of course, I have inserted so many off-topics. Don’t feel annoying. Blogging is still not easy, and let the blogger nag a few words
Back to the topic, the above result +1, becomes 10000 4 =4 5 , then 4 1 +4 2 +4 3 +4 4 = (4 5 -1)/3, then, the same operation
33333……(n) 4 If there are n 3s, is the final geometric sequence result (4 n+1 -1)/3
Small summary
The general term is a n = 2 n using the binary system, the general term is a n = 4 n , using the quaternary system. I personally feel that it is possible to derive the sum of the first n terms of the geometric sequence in an infinite system. However, 10 I think it’s not very suitable for the base system. At this time, you can still use this method by factoring
Conclusion
In fact, many things in mathematics require everyone’s insights to truly understand them. Whether you have learned mathematics well or not, your cold scores can’t prove it for you. There are many examples of this around me, and I usually talk to others about high math problems. It’s also very clear. The exam is not high enough. What's the solution? You can’t remember that time period, just can’t remember..., once time passes, game over...
Publicity
As for me, I am a junior. Currently, I mainly explore the direction of big data development, and I have also contacted some big data development tools such as hadoop, spark, hive, etc., and I will successively upload some blog posts about big data in the future. This is a wave of your own contact information, QQ + WeChat: 928689419, just add me to verify the big data
Easter eggs
The above derivation is the case of proportional q>1, and the case of adding q<1 is listed as a n =(1/4) n , how to find it?
You are welcome to leave comments and give your own solutions. I have not given a proof here, but I believe that my friends here can definitely make it.