0

Hi I'm following a course on mathematical induction, and I found something that doesn't quite make sense to me

The professor gave os this statement: $$3+7+11+\cdots+(4n-1) = n(2n+1)$$

then he goes straight on to test this with $n=1$, and goes straight to $$3 = n(2n+1)$$

and in the case where $n=2$, he simply includes two numbers in the sets $$3 + 7= n(2n+1)$$ So here n is simply the amount of numbers that are included on the left side of the equal sign. But how is this the case? it seems to me that $3+7+11+\cdots+(4n-1)$ should be $3+7+11*(4n-1)$ which would just output some number

What am I not understanding?

Souza
  • 889
s3j80
  • 1
  • 2
  • For $n=1$ we have $n(2n+1)=3$, so far so good? – Dietrich Burde Apr 21 '20 at 12:48
  • 2
    It is not $3+7+11\times (4n-1)$, it is $3+7+11+\dots+(4n-1)$. This is just shorthand for writing $\sum\limits_{k=1}^n(4k-1)$. That is to say when $n$ is $1$ that would be $3$. When $n$ is $2$ it would be $3+7$. When $n$ is $3$ it would be $3+7+11$. When $n$ is $4$ it would be $3+7+11+15$. In general, for $n$ in general you have the sum of the first $n$ positive numbers which are one less than a multiple of four. – JMoravitz Apr 21 '20 at 12:48
  • It must be shown here that $\sum_{k=1}^n(4k-1)=n(2n+1)$ for every positive integer $n$. Is that clear to you? – drhab Apr 21 '20 at 12:49
  • 1
    The dots there simply mean that the pattern is the same and can be filled in by your own mind, but keeping in mind that for small enough $n$, the pattern can end early, and even earlier than what you might expect from the terms visible. – JMoravitz Apr 21 '20 at 12:50
  • dietrich, not really, it shouldnt it be if n=1 then 1(2*1+1)? – s3j80 Apr 21 '20 at 12:54
  • @s3j80 there is no difference. If $\color{blue}{n}=\color{blue}{1}$ then $\color{blue}{n}(2\cdot \color{blue}{n}+1)$ is the same as $\color{blue}{1}(2\cdot \color{blue}{1}+1)$, both of which equal $3$. – JMoravitz Apr 21 '20 at 12:55
  • and if (2n+1) is the way incrementation is made, then shouldnt the incrementation of the three first elements depend on n? – s3j80 Apr 21 '20 at 12:56
  • Let's back up a bit then... Do you understand that when I write $1+2+3+\dots+10$ that this is shorthand for $1+2+3+4+5+6+7+8+9+10$? Do you understand then what $1+2+3+\dots+10000$ means and why writing it with dots is preferable to writing out all ten-thousand terms in the sum? – JMoravitz Apr 21 '20 at 12:58
  • yes yes, completely. does this mean that (4n-1) is simply the end of the sequence? – s3j80 Apr 21 '20 at 12:59
  • Yes... that's all that is going on here. $3+7+11+\dots+(4n-1)$ is a sum with $n$ summands (the individual terms being summed) whose first term is $3$, whose terms increase by $4$ each time, and whose last term is $4n-1$. (Noting that in the case that $n=1$ the first term and the last term are one and the same). – JMoravitz Apr 21 '20 at 13:01
  • this makes sense, I can see the pattern in how the numbers get larger, but i dont see why if n=1, then we act like 7 and 11 doesnt exist. wouldnt it just mean the set would end in 3? – s3j80 Apr 21 '20 at 13:04
  • When $n=1$ we have the first term is $3$, the last term is $(4\cdot 1 -1)$ which is also $3$... and so yes, when $n=1$ there being only one term in the sum that would correspond to the sum $3$ with nothing else added to it. Yes, that means we "pretend like 7 and 11 don't exist" because they don't. They were only written in the original expression so that the pattern is easier to spot that the terms grow by $4$ each time. You start with $3$, and until you are done, you add the next term one less than a multiple of four until you reach $4n-1$ which happens right away for $n=1$. – JMoravitz Apr 21 '20 at 13:08

0 Answers0