Find a formula for $\sum_{i=1}^n (2i-1)^2 = 1^2+3^2+....+(2n-1)^2$

Question

Consider the sum $$\sum_{i=1}^n (2i-1)^2 = 1^2+3^2+...+(2n-1)^2.$$

I want to find a closed formula for this sum, however I'm not sure how to do this. I don't mind if you don't give me the answer but it would be much appreciated. I would rather have a link or anything that helps me understand to get to the answer.

EDIT: I Found this question in a calculus book so I don't really know which tag it should be.

The statement itself is wrong to begin with. $\displaystyle \sum_{i=1}^n (2i-1)^2 = 1^2+3^2+5^2+\cdots+(2n-1)^2$. — DHMO, Apr 12 '17 at 14:15
Also, it should end on $(2n-1)^2$ and not $(2i-1)^2$ which makes no sense. — Mathematician 42, Apr 12 '17 at 14:20
What @DHMO is saying is that $i$ is defined only within the sum. Therefore, it can't appear on the RHS. Instead, the RHS should have only the variable $n$. — Michael Burr, Apr 12 '17 at 14:20
Everyone cared about the last term. Only @samjoe noticed that the middle terms are also wrong. — DHMO, Apr 12 '17 at 14:21
@DHMO: I think you should edit the question, the mistakes are still there after a couple of edits. — Mathematician 42, Apr 12 '17 at 14:25
Just realized all the mistakes I made ugh. This is the result of lack of sleep and too much coffee. Ty for editing the mistakes. — Tom Harry, Apr 12 '17 at 18:14
At least connect this to an old post that is a target of several duplicate posts. — Lee David Chung Lin, Jan 11 '19 at 07:36

Mathematician 42 · Answer 1 · 2017-04-12T14:41:02.870

3

Hint: $\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}{6}$ and $\sum_{i=1}^ni=\frac{n(n+1)}{2}$. And last but not least $$(2i-1)^2=4i^2-4i+1.$$

Edit: Let's prove that $\sum_{i=1}^ni=\frac{n(n+1)}{2}$. We proceed by induction on $n$. If $n=1$ the statement is trivial. Now suppose the statement holds for $n\geq 1$. Then \begin{eqnarray}\sum_{i=1}^{n+1}i&=&\sum_{i=1}^ni+(n+1)\\ &=&\frac{n(n+1)}{2}+(n+1)\\ &=& (n+1)(\frac{n}{2}+1)\\ &=& \frac{(n+1)(n+2)}{2}.\end{eqnarray} Here we used the induction hypothesis in the second equation. This proves the statement by induction. You can prove the other formula in a similar fashion.

edited Apr 12 '17 at 14:41

answered Apr 12 '17 at 14:19

Mathematician 42

12,591

$\frac{4n(n+1)(2n+1)}{6} - \frac{4n(n+1)}{2} + n$ will it look something like this? and where can i learn about these formulas etc. – Tom Harry Apr 12 '17 at 14:29
Yes, if you work it out you should get $\frac{1}{3}n(4n^2-1).$ I'll edit my answer and show you how to prove one the formulas. – Mathematician 42 Apr 12 '17 at 14:34

hamam_Abdallah · Answer 2 · 2017-04-13T17:39:30.757

1

hint

We have

$$1^2+3^2+5^2+... (2n-1)^2=$$

$=\sum $ odd$^2$=$\sum$ all$^2 $-$\sum $even$^2=$

$$\sum_{k=1}^{2n} k^2-(2^2+4^2+...4n^2)= $$

$$\sum_{k=1}^{2n}k^2-4\sum_{k=1}^n k^2=$$

$$\boxed {\color {green}{\frac {n(4n^2-1)}{3}}}$$

for $n=2$, we have $10 $ , for $n=3$, we find $35$ and for $n=4$, it is $84$.

edited Apr 13 '17 at 17:39

answered Apr 12 '17 at 14:19

hamam_Abdallah

62,951

i don't understand how you got to $\sum_1^{2n} k^2-(2^2+4^2+...4n^2)=$ – Tom Harry Apr 12 '17 at 14:38
plus what does $\sum_1^{2n} $ this even mean. How can it be 2n isnt it essentially just n – Tom Harry Apr 12 '17 at 14:40
@TomHarry I edited it . now it is clear. – hamam_Abdallah Apr 12 '17 at 14:43
doesn't sigma all^2 = $\sum_{k=1}^{n} k^2$ and if so why do u have $\sum_{k=1}^{2n} k^2$ – Tom Harry Apr 13 '17 at 16:05

score 0 · Answer 3 · answered Apr 12 '17 at 14:22

Hint: Write $f(n) = \sum_{i=1}^n (2i-1)^2$. You can start by hypothesizing that the sum of squares is a cubic $f(n) = an^3 + bn^2 + cn + d$ (as a sort of discrete analogy with the integration formula $\int x^2\, dx = x^3/3 + C$). Then $(2n+1)^2 = f(n+1) - f(n) = a (3n^2 + 3n + 1) + b(2n + 1) + c$, which gets you the constants $a, b, c$, and you can find $d$ from $f(1) = 1$. Verifying this formula is an easy proof by induction.

Hypergeometricx · Answer 4 · 2017-04-13T03:36:44.103

$$\begin{align} \sum_{i=1}^n (2i-1)^2 &=\sum_{i=1}^n \binom {2i-1}2+\binom {2i}2&&\text(*)\\ &=\sum_{r=0}^{2n}\binom r2\color{lightgrey}{=\sum_{r=2}^{2n}\binom r2}\\ &=\color{red}{\binom {2n+1}3}&& \text(**)\\ &\color{lightgrey}{=\frac {(2n+1)(2n)(2n-1)}6}\\ &\color{lightgrey}{=\frac 13 n(2n-1)(2n+1)}\\ &\color{lightgrey}{=\frac 13 n(4n^2-1)} \end{align}$$

*Note that $\;\;\;R^2=\binom R2+\binom {R+1}2$.
**Using the Hockey-stick identity

mocooJ · Answer 5 · 2023-07-21T04:40:15.397

There has grneral method,no longer requires various skills.See Redefining the Shape of Numbers and Three Forms of Calculation
define:

$$SUM(N,[{K_1},{K_2}...{K_M}]:D,[1,2...M]) = \sum\limits_{i = 0}^{N - 1} {({K_1} + nD)} ({K_2} + nD)...({K_M} + nD)$$ then $$SUM(N,[{K_1},{K_2}...{K_M}]:D,[1,2...M]) = \sum\limits_{g = 0}^M {{H_1}} (g)\left( {_{g + 1}^N} \right)$$ Use an auxiliary form for calculating H₁(g). $$SET - T = \{ {T_1},{T_2}....{T_M}\} = \{ 1,2...M\} ,SET - K = \{ {K_1},{K_2}....{K_M}\} $$ $$({K_1} + {T_1})({K_2} + {T_2})...({K_M} + {T_M}) = \sum {\coprod\limits_{i = 1}^M {{X_i}} } ,{X_i} = {T_i}\,or\,\,{K_i}$$ Define: $$X(T) = count{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} of{\kern 1pt} {\kern 1pt} {\kern 1pt} {X_i} \in SET-T $$ $${X_{T - 1}} = count{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} of\{ {X_1},{X_2}...{X_{i - 1}}\} \in \{ {T_1},{T_2}...{T_{i - 1}}\} $$ $${X_{K - 1}} = count{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} of\{ {X_1},{X_2}...{X_{i - 1}}\} \in \{ {K_1},{K_2}...{K_{i - 1}}\} $$ $${X_{T - 1}} + {X_{K - 1}} = i - 1$$ Each ∏X_i corresponds to one expression in the sum. $${H_1}(g) = \sum\limits_{X(T) = g} {\prod\limits_{i = 1}^M {{B_i}} } .If\,{X_i} = {T_i}{\kern 1pt} {\kern 1pt} {\kern 1pt} then{\kern 1pt} {\kern 1pt} {B_i} = ({T_i} - {X_{K - 1}})D{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} else{\kern 1pt} {\kern 1pt} {\kern 1pt} {B_i} = {K_i} + {X_{T - 1}}D$$

In this question:

$${\text{SUM(n ,[1,1]:2,[1,2])}} = {H_1}(2)\left( {_3^n} \right) + {H_1}(1)\left( {_2^n} \right) + {H_1}(0)\left( {_1^n} \right)$$ $$Form:({K_1} + {T_1})({K_2} + {T_2}) = {K_1}{K_2} + {K_1}{T_2} + {T_1}{K_2} + {T_1}{T_2},don't{\kern 1pt} {\kern 1pt} swap{\kern 1pt} {\kern 1pt} {\kern 1pt} factors$$

$$\sum\limits_{X(T) = 1} {\prod {{B_i}} } = {T_1}{K_2} + {K_1}{T_2},{H_1}(1) = 1 \times D \times (1 + D \times {X_{T - 1}}) + 1 \times (2 - {X_{{\text{K}} - 1}}) \times {\text{D}}$$ $$\to {H_1}(1) = 1 \times 2 \times (1 + 2) + 1 \times (2 - 1) \times 2 = 8$$

$${H_1}(0) = 1 \times 1 = 1,{H_1}(2) = 2!{2^2} = 8$$ $$Answer = 8\left( {_3^n} \right) + 8\left( {_2^n} \right) + \left( {_1^n} \right) = \frac{{n(4{n^2} - 1)}}{3}$$

While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review — PC1, Jul 21 '23 at 02:53

Find a formula for $\sum_{i=1}^n (2i-1)^2 = 1^2+3^2+....+(2n-1)^2$

5 Answers5

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