# The sum of the $n$ smallest odd numbers is equal to $n^2$ [duplicate]

if given a positive integer $n$ is the sum of the $n$ smallest odd numbers equal to $n^2$

• Base case: Prove/show/verify for $n=1$. Induction hypothesis: assume true for $n=k$. Step 3: Prove true for $n=k+1$. – paw88789 Aug 26 '14 at 23:13
• It looks as if you want to show this: The sum of the $n$ smallest odd positive integers is equal tn^2$. Induction is one way to do it. – André Nicolas Aug 26 '14 at 23:13 • @Andre N: Sum of$n$smallest odd positive integers, I think. – paw88789 Aug 26 '14 at 23:15 • Yes, forgot to write odd. You want to show that if the sum of the odds up to$2k-1$is$k^2$, then the sum of the odds up to$2(k+1)-1$is equal to$(k+1)^2$. This is easy, because$k^2+(2k+1)=(k+1)^2$. – André Nicolas Aug 26 '14 at 23:17 • A few suggestions: (1) try to search before asking. Most standard induction questions have been asked and answered before. (2) Write a descriptive title for your question. This is for your own benefit: the system will suggest related posts based on title. (3) Use LaTeX. – user147263 Aug 26 '14 at 23:25 ## 3 Answers You want to show$1+3+\cdots+(2n-1)=n^2$. Note that$1+2+\cdots+2n=n(2n+1)$, and$2+4+\cdots+2n=2(1+2+\cdots+n)=n(n+1)$. Can you prove these two formulas? Your result now follows by subtraction. • @yeyee6512y: Let$S=1+2+\cdots+n$=$n+(n-1)+\cdots+1$. So,$2S=n(n+1)$. Or, prove this statement by induction. – voldemort Aug 26 '14 at 23:33 • @yeyee6512y: Check that last step-it isn't correct.$k^2+2k-1+1=k^2+2k=k(k+2)$. – voldemort Aug 26 '14 at 23:38 • @yeyee6512y: sorry- edited . – voldemort Aug 26 '14 at 23:41 • by factoring :). – voldemort Aug 26 '14 at 23:43 • It's not- you should check your computation :). All I am suggesting is that write$1+2+..n$in one row, and write$n+(n-1)+..1$in the next row, and add term by term- you will see a magic happen :). – voldemort Aug 26 '14 at 23:46 My favorite, the proof without words: • Indeed. That is pretty sweet. – Graham Kemp Aug 26 '14 at 23:42 • @GrahamKemp Thanks; does that get me an upvote? ;) – Ahaan S. Rungta Aug 26 '14 at 23:43 First we observe that the pattern seems to hold.$1^2 = 1, 2^2 = 1+3 , 3^2 = 1+3+5, \ldots$Then we notice that this is:$2^2 = 1^2 + 3, 3^2=2^2 + 5, \ldots$and we predict that: $$(n+1)^2 =\mathop{(n)^2 + \underbrace{(2(n+1)-1)}}_{\text{the }(n+1)^{th}\text{ odd number}}$$ For a proof by induction,$\require{cancel}\cancelto{\text{you}}{\text{we}}$need to demonstrate that this is so for all positive integer$n$. (Having already established the base case:$n=1: 1^2=1) More formally: The steps in proof by induction are: \begin{align} & \mathsf P(1) & \text{base case}\\ & \forall n\in\mathbb{Z}^+ : \mathsf P(n)\to \mathsf P(n+1) & \text{iterative case} \\ \therefore \quad & \forall n\in\mathbb{Z}^+ : \mathsf P(n) & \text{by induction}\end{align} Our premise is:\mathsf P(n):= (n^2=\sum_{k=1}^n (2n-1))$Base case:$1^2= (2(1)-1) \quad\color{green}{\checkmark}$Iterative case:$\Bigl(n^2 = \sum_{k=1}^n (2n-1))\implies \underbrace{(n+1)^2 = n^2 + (2(n+1)-1)}_{\text{show this is true for all }n\in\mathbb Z ^+}\Bigr) \quad \color{green}{?}$Notes The$k^{th}$odd number is$2k-1$. As in:$1=2(1)-1, 3=2(2)-1, 5=2(3)-1, \ldots$. Then the sum of the first$n$odd numbers is:$a_n:=\sum_{k=1}^n (2k-1)\begin{align}a_n &:= 1+3+5+\cdots+(2n-1)\end{align} It follows that the sum of the firstn+1$odd numbers must be:$a_{n+1} = \sum_{k=1}^{n+1} (2k-1)\begin{align}a_{n+1}&:= 1+3+5+\cdots+(2 (n)-1)+(2 (n+1)-1)\end{align} Hence by subtraction it is that:a_{n+1} - a_{n} = (2(n+1)-1)$Now if it is that:$a_n=(n)^2$then it would also be that:$a_{n+1} = (n+1)^2$. Thusly our induction step needs to show:$(n+1)^2 = (n)^2 + (2(n+1)-1)\$

• @yeyee6512y See the added notes. (Also, corrected a typo.) – Graham Kemp Aug 27 '14 at 0:46
• @yeyee6512y Which step don't you follow? – Graham Kemp Aug 27 '14 at 0:50