Picturing 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 $51 + 53 + 55 + ... + 149 + 151 + 153 = 77^2 - 25^2 = 5304$

Ian from Colton Primary School and Hannah from Thorner C of E Primary School also completed particularly nice solutions, but we don't have space to show them here .