Quadratic patterns

We have received many excellent solutions for these problems. Well done to all of you who included numerical, diagrammatic and algebraic representations and descriptions in words for these problems. Particularly well done to Diana from Dulwich College Beijing in China, Sameera from Ripon Grammar School, Amrit from Hymers College, Chloe, Sophia and Shreya from North London Collegiate School, Maddy and Grace from the Stephen Perse Foundation and Peter from Durham Johnson School in the UK, Aarav, Ruhi, Sehar, Nikhil, Vishnuvardhan, Ahana, Ananthajith, Dhruv, Vihaan, Arya, Aaradhya and Vishnu from Ganit Kreeda, Vichar Vatika, India, Shaunak also from India, Miriam from IES Maximo Laguna and Marc and Yang from King's College Alicante in Spain, Radi and Camilla and Viktor and Matija from European School Varese in Italy, Ziqian from Harrow International School in Hong Kong,  Rubayat, Hondfa, Jacob and Nathan from Greenacre Public School in Australia, and Ryan from Dulwich College Seoul in Korea.

1.

Amrit from Hymers College in the UK, Radi and Camilla and Viktor and Matija from European School Varese in Italy, Rubayat, Hondfa, Jacob and Nathan from Greenacre Public School in Australia, Ryan from Dulwich College Seoul in Korea and Peter from Durham Johnson School in the UK sent in good work on this problem.

Viktor and Matija spotted a numerical pattern and summarised it in words:
 

$2 \times 3 + 3 =9$

$5 \times 6 + 6 =36$

$4 \times 5 + 5 =25 $

$9 \times 10 + 10 =100$

If you multiply the two consecutive numbers, which differ by one, then add the bigger number to that you're always going to get the square of the bigger number.

Ryan showed numerically why this works:

$$\begin{align}2 \times 3 + 3 = 9   &=  3 + 3 + 3 = 3 \times 3 = 9\\

5 \times 6 + 6 = 36  &=  6 + 6 + 6 + 6 + 6 + 6 = 6 \times 6 = 36\\

4 \times 5 + 5 = 25  &=  5 + 5 + 5 + 5 + 5 = 5 \times 5 = 25\\

9 \times 10 + 10 = 100 &=  10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10+ 10 = 10 \times 10 = 100\end{align}$$

 

Amrit, Sehar, Nikhil, Vishnuvardhan, Ahana, Ananthajith, Dhruv, Vihaan, Arya, Aaradhya and Vishnu gave algebraic proofs. This is Amrit's algebraic proof:
 

Represented algebraically, this is $n(n+1)+(n+1)$

Factorising, $(n+1)(n+1)$

And we see this is $(n+1)^2$

Sehar, Nikhil, Vishnuvardhan, Ahana, Ananthajith, Dhruv, Vihaan, Arya, Aaradhya and Vishnu combined algebra and words to explain this result:

$n(n+1)$ means $(n+1)$ added $n$ times.

$n(n+1)+n+1$ means $(n+1)$ added $(n+1)$ times.
So, $n(n+1) +n+1 = (n+1)(n+1)=(n+1)^2$.

They then represented this in a diagram. The rectangles with black sides show the left hand side of their equation and the red square shows the right hand side.

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Radi and Camilla gave a diagram which represented the equation using similar ideas:

 

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2.

This is Radi and Camilla's work for the second part of the problem:

 

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Radi and Camilla used $a, a+1$ and $a+2$ to represent three consecutive numbers. Diana and Sameera used a slightly different algebraic representation. Here's Sameera's working. Can you see how this represents the three consecutive numbers?

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

Sameera also sent in this image of her diagrams that explained the result. Unfortunately the picture is a bit blurred, but can you use her diagrams to work out what her explanation might be?

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3.

Viktor and Matija spotted the numerical pattern for the third part of the problem:

$3\times3-1\times1=8$

$8\times8-6\times6=28$

$7\times7-5\times5=24$

$10\times10-8\times8=36$

Amrit put it in words:

If you take two numbers with a difference of 2 and subtract the square of the smaller one from the square of the bigger one, the result is four times the mean of the two numbers.

Aarav noticed something about how the values changed and put this into words:

I noticed that by each increasing pattern the answer of equation increases by 4.

Can you connect these two ways of describing the pattern with the other representations?

Tiago and Finn and Shaunak proved it algebraically. Here's Tiago and Finn's proof:

$$\begin{align}(n\times n)-(n-2)\times(n-2)=&n^2-(n-2)\times(n-2)\\

=&n^2-(n^2-4n+4)\\

=&n^2-n^2+4n-4\\

=&4(n-1)\end{align} $$

 

Shaunak used $n$ and $n+2$ as her two numbers that differ by 2, which gave a slightly different algebraic argument: 

$(n+2)^2-n^2=n^2+4n+4-n^2=4n+4=4(n+1)$.

Peter thought about it differently:

The answer is always the sum of the digits [multiplied by 2]. $a\times a-b\times b=2b+2a$ if $a-b=2$.

Marc and Yang called the numbers $x$ and $y$ and noticed that

$x^2 - y^2 = (x+y)(x-y)$

Amrit's proof used Marc and Yang's idea to prove Peter's representation in words.

$(x+2)^2-x^2=(x+2+x)(x+2-x)$, which simplifies to $(2x+2)(2)$

 

Tiago and Finn's diagram representation looked like this:

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Like her algebra, Shaunak's diagram was slightly different. Can you relate the diagram to Shaunak's algbera and connect her argument with Tiago and Finn's diagram?

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4.

Peter from Durham Johnson School in the UK noticed that the expressions all simplify to 2:

$n(n+1)−(n−1)(n+2)=2$

$(n+1)(n+2)−n(n+3)=2$

$(n−3)(n−2)−(n−4)(n−1)=2$

The following observations can be made:

  "If you choose four consecutive numbers and subtract from the product of the two middle ones the product of the other two, then you always get $2$."

   General expression: $n(n+1) - (n-1)(n+2) = (n^2 + n) - (n^2 + n - 2) = 2$.

   It follows that $(n+1)(n+2) - n(n+3)$ and  $(n-3)(n-2)-(n-4)(n-1)$ are always $2$.

These diagrams represent the process. The shaded area in the right hand diagram needs to be 'subtracted' from the green area.

 

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This leaves a gap of 2 squares:

      

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5.

Maddy and Grace from the Stephen Perse Foundation in the UK continued the numerical pattern:

$2 \times 4 + 1 = 9$

$4 \times 6 + 1 = 25$

$5 \times 7 + 1 = 36$

$9 \times11 + 1 = 100$

$10 \times 12 + 1 = 121$

$20 \times 22 + 1 = 441$

Rubayat, Hondfa, Jacob and Nathan from Greenacre Public School in Australia noticed that

The pattern is what Charlie quoted,"If you multiply two numbers that differ by 2, and then add one, the answer is always the square of the number between them!" $3 \times 5 + 1= 16$ or $4^2$. $5 \times 7 + 1 =36$ or $6^2$.

Peter's proof was different to Charlie's:

$(n-1)(n+1)= n^2-1$, so $(n-1)(n+1)+1=n^2$

Chloe, Sophia and Shreya from North London Collegiate School made a clear diagram with a good explanation:

 

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Ziqian included a similar diagram, but explained the result as a rearrangement of the pieces:

In the diagram we can also see that by moving the pieces and adding 1 we get the number we started from $=x^2$.

6.

Rubayat, Hondfa, Jacob and Nathan represented the problem numerically and in words:

The result equals the middle number squared, then times by $2$. For example: $1,2,3.$ $1 \times 2=2,$ $2 \times 3=6$ therefore $6 +2 = 8$ [which is $2^2$ times $2$].

Another example is $50,51$ and $52.$ $50 \times 51= 2550,$ $51 \times 52= 2652.$ Therefore

$2550 + 2652=5202.$ $51$ squared then times $2$ equals $5202.$

Amrit proved it algebraically:

Represented algebraically, this is

$$x(x+1)+(x+1)(x+2)$$Factorising,$$(x+1)(x+x+2)$$Simplifying,$$(x+1)(2x+2)$$Factorising,$$2(x+1)(x+1)$$Simplifying,$$2(x+1)^2$$

Clare made a diagram:

 

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Sehar, Nikhil, Vishnuvardhan, Ahana, Ananthajith, Dhruv, Vihaan, Arya, Aaradhya and Vishnu used a similar diagram,  but like Sameera, they had used $n-1, n$ and $n+1$ as their three consecutive numbers so their diagram is slightly different:

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 We had so many good solutions to this problem. Well done again to everyone who submitted solutions!