Picturing square numbers

We've received lots of good solutions to this problem. Well done to all of you who sent in correct solutions. A lot of you spotted the connection with square numbers. 

 

Hannah from Millom School in Cumbria sent in a nicely articulated solution:

The diagram shows that the sum of the first $4$ odd numbers is $16$ because there are $4$ rows and $4$ columns of counters. For the sum of the first $20$ odd numbers there are $20$ rows and $20$ columns. So if I do $30 \times 30$ (or $30$ squared) I get an answer of $900$. For the sum of the first $60$ odd numbers there are $60$ rows and $60$ columns. So if I do $60 \times 60$ (or $60$ squared) I get an answer of $3600$.

If you want the sum of the first $n$ odd numbers the answer would be $n$ squared.

I worked out that $153$ is the $77$th odd number. I did this by adding one (to get $154$) and then divided the answer by $2$.

The sum of the first $77$ odd numbers is $77\times77$ which is $5929$.

To find $51 + 53+ 55+\ldots+ 149 + 151 + 153$ I used the answer from the previous question which was $5929$.

As we were starting at $51$ this time and not $1$, I needed to find the sum of all the odd numbers from $1$ up to $49$. I found that $49$ is the $25$th odd number (by adding $1$ to $49$ and then dividing the answer by $2$) So the sum of the odd numbers from $1$ to $49$ is $25$ squared which is $625$.

Finally I took $625$ away from $5929$ to give an answer of $5304$.

David decided to use algebra to explain his thinking:

The sum of the first $30$ odd numbers $= 30^2 = 900$.

The sum of the first $60$ odd numbers $= 60^2 = 3600$

Quick Method: The sum of the first $n$ odd numbers $= n^2$

What is the sum of $1 + 3 + \ldots + 149 + 151 + 153$?

The formula for odd numbers is $2n-1$

We have: $2n-1 = 153$

$2n = 154$

$n = 77$

So $153$ is the $77$th odd number. The sum of the first $77$ odd numbers $= 77^2 = 5929$. Therefore, the sum of $1 + 3 + \ldots+ 149 + 151 + 153 = 5929$

What is the value of $51 + 53 + 55 + \ldots+ 149 + 151 + 153$?

The answer is the sum of ($1 + 3 + \ldots + 149 + 151 + 153$ - which is already worked out) minus the sum of ($1 + 3 + \ldots +49$)

$49$ is the $25$th odd number (as $2n-1 = 49 \Rightarrow 2n = 50$, so $n = 25$)

Therefore the value of $51 + 53 + 55 + ... + 149 + 151 + 153 = 77^2 - 25^2 = 5304$

Shaunak from India also used algebra, but specifically used the method of proof by induction to prove the connection with square numbers. You can see this approach here.

The method I found is if we have to add up to the first $k$ odd numbers, the sum is $k^2$
Proof: I will prove by induction. 
Base Case: $k = 1$ 
$1 = 1^2$ 
$1 = 1$ 
Induction Hypothesis: 
Assume $1 + 3 + 5 + \cdots +k^{th} \,\rm{odd \,number} = k^2$
To prove: $1 + 3 + 5 + \cdots +k^{th} \,\rm{odd \, number} + (k+1)^{th} \, \rm{odd \,number}  = (k + 1)^2$ 
Induction step: If n is the $k^{th}$ odd number, then $n = 2k – 1$. 
Then $1 + 3 + 5 + … + 2k – 1 = k^2$ 
Add $2(k + 1) – 1 = 2k + 1$ to both sides. 
$1 + 3 + 5 +\cdots + 2k – 1 + 2k + 1 = k^2 + 2k + 1$ 
$k^2+ 2k + 1 = (k + 1)^2$
Therefore $1 + 3 + 5 + \cdots + 2k – 1 + 2k + 1 = (k + 1)^2$. Hence proved.

We asked you whether $3249$ could be made by adding odd numbers in this way. We had lots of solutions for this that used the connection with square numbers. Turker from the Mark Twain IS239 Intermediate School for the Gifted and Talented in US gave this explanation:

You can make $3249$ by adding odd numbers

$n^2=3249$

$\sqrt{n^2}=\sqrt{3249}$

$n=\sqrt{3249}$

$n=57$.

The first 57 odd numbers will add up to $3249$.

Like David, a lot of you noticed that $51 + 53 + 55 + \cdots+ 149 + 151 + 153$ is the difference of two sums of odd numbers and you could then use your previous answer to $1 + 3 + 5 + \cdots+ 149 + 151 + 153$. Ziqian from Harrow International School Hong Kong represented this with a diagram. As Ziqian said, 

This question is a bit trickier, for it doesn’t start from 1, but 51. So we have to imagine it as a squared number subtracted by a square number.

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Kanaa from Ganit Kreeda came up with an alternative approach. She subtracted 50 from each of the 52 terms in the sum.

$51 + 53 + 55 + \cdots+ 149 + 151 + 153$

$-50 -50 -50 -\cdots -50 -50 -50$

She then worked out the value of $1+3+5+ \cdots+99+101+103$ using her previous method and added the on the $52$ lots of $50$.

$1+3+5+ \cdots+99+101+103$

$=52 \times 52$

$=2704$

$52\times 50 = 2600$

Final answer $=2704+2600=5304.$

We had lots of solutions to the last two sums, so well done to everyone who sent us a solution for these.  Kanaa used her same method to find the fourth sum by subtracting $1$ from each term. This turned the sum into the previously solved $1+3+5 + \cdots +151+153$. Therefore her final answer was $5929+77=6006.$ Brilliant work Kanaa for turning a new problem into one you'd already solved!  

 Sanat, aslo from Ganit Kreeda sent a video sharing his approach.

Many of you also spotted that you can work out the sums in this problem using the method in Slick summing. Well done to Aarna from Wallington High School for Girls, Salaar from Durham Johnston Comprehensive School, Jiaqi from Christ Church CE Primary School, Arjun, Romany and Oliver from Pate's Grammar School and Siena from Guildford High School, all in UK, Enzo from Harrow International School in Hong Kong, and Amay and Ananya from Ganit Kreeda in India and Shaunak also in India for making this connection! 

Finally, a group of students from Ganit Keeda explored this problem further using similar diagrams to the one we started with. Here's a taster of the problems and the approach taken by Varun, Thejasvi and Prapthi: 

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So the total number of dots $=n^2+3n$ where $n=$ number of colours. 
 

We'd love to hear if these extensions of the problem have given you any ideas for connecting $2+4+ \cdots +150+152+154$ and $2+6+ \cdots+298+302+306$ with the starting diagram. Please email us if you've got any ideas about this.