Slick Summing

We've received lots of good solutions to this problem. Thanks to Lasse (from St. James Senior Boys' School), Fred (from Wembley High Technology College), Nishil (from Cambridge School), Rishi (from St. Matthew's Primary School), Zachary (from Sonoran Science Academy), Kaushal (from Overseas Family School), Rukkshana and Annie (from Fern Avenue Public School), Julia and Grace (from Inter-Lakes), Julian (from British School Manila), Poppy (from Queen Elizabeth's Grammar School in Horncastle), Ci Hui (from Queensland Academy for Science Mathematics and Technology, Australia), Shaunak (from India) and Qiun (from China) for all the great answers. 

Poppy used a diagram to represent Charlie's method as part of working on this problem:

Image

Here's Zachary's solution to the first thee sums in the problem:

First, let's work out a solution to 1+2+3+...+10.

When we pair the first and last term, 2nd and 9th term, 3rd and 8th term, ..., we get sums of 11. We can list the pairs:

• 1 and 10
• 2 and 9
• 3 and 8
• 4 and 7
• 5 and 6

There are 5 pairs of sums that add up to 11, and so we get a sum of 5*11 = 55.

Now, we need to find 1+2+3+...+20. Now listing out the terms would be a bad idea, so let's think about how many terms that was in the sum of 1-10. There were 5 pairs in 1-10. So, if the same rules apply, then there would be 10 pairs in the sum of 1-20. Now, we find the sum by adding the first and the last terms of the sequence. 1+20= 21. We have 10 pairs of 21, and the sum is 10*21= 210.

We will do the same thing for 1-100. 100/2= 50 pairs. 1+100= 101. 50 pairs of 101 is 50*101= 5050.

Now, we have the sum of 40-100. This is harder to solve, because the first term isn't 1. So, let's start by finding the number of terms. First, we must realize that in the set of numbers from n to m, there are m-n+1 terms. So here we have the numbers from 40 to 100, so there are 61 terms. We use the same method, even though there are not an even number of terms. 61/2=30.5. 100+40= 140. 140*30.5= 4270.

Brilliant! 

Shaunak offered an alternative to this method for the sum fo 40-100. Their calculation was 

$5050-39 \times \dfrac{40}{2}$

How would you explain why this will give the same answer as Zachary's method? 

Lasse, Ci Hui and Shaunak pointed out that sums like this are called 'arithmetic series' - well spotted. Kaushal proved a general formula for us:

The sum 1 + 2 + 3 + 4 ... + n-1 + n can be rearranged:

(1 + n) + (2 + (n - 1)) + (3 + (n - 2)) +...

where each term is in the form (x + (n - (x-1)). This simplifies to:

(1 + n) + (1 + n) + (1 + n) +...

The number of ''(1 + n)''s that are added is n/2 since (x + n - (x-1)) is always n + 1. Therefore,

$1 + 2 + 3 + 4 +\dots + n-1 + n =   \dfrac{n^2 + n}{2}$.

Julian gave us the following answers:

• 1+3+5+...+17+19 is (20*10)/2 which is 100
• 2+4+6+...+18+20 is (22*10)/2 which is 110
• 42+44+46...+98+100 is (142*30)/2 which is 2130

Great - but we then asked where these come from. 

Aarna, from Wallington High School for Girls, UK, gave a solution with further details for the sum  $42 + 44+ \cdots + 98+100.$ 

First she found the sums of pairs of numbers:

$42+100=142$ 

$44+98=142$ 

$46+96=142$ 

$\vdots$ 

$98+44=142$ 

$100+42=142$

Aarna then worked out the number of these pairs by counting the number of even numbers from 42 to 100. She found that there were five even numbers in each group of ten numbers such as $42-50$ and there were six of these groups from $42-100$.  As she explains,

The total number of even numbers  $42 - 100 =6\times 5=30$.

$30=$ number of pairs

$30 \times 142=4260$

$4260 \div 2 = 2130$

So: $42+44+46+\cdots +98+100=2130.$

Other approaches involved more algebra. Ci Hui showed how the answers James had found came from the formula for the sum of an arithmetic series. If the first term is $a$ and the common difference is $d$, then the sum of the first $n$ terms is $\dfrac{n}{2} (a + a +(n-1)d)$ or $\dfrac{n}{2} (2a +(n-1)d)$. Can you connect this formula to Charlie's method?

Although this formula can be used for all these calculations, Shaunak found specific formulae for the sums of the first $k$ odd numbers and the first $k$ even numbers. As she explains:

$n$ is the last term of the progression. For AP (arithmetic progression) with a difference of 2, containing odd numbers, the formula is $\left(\dfrac{n + 1}{2}\right)^2$:

$1 + 3 + 5 + … + 2k – 1$

$2k – 1 + … + 5 + 3 + 1$

$\dfrac{(2k + 2k + 2k + … + 2k + 2k)}{2} = \dfrac{k(2k)}{2} = \dfrac{2k^2}{2} = k^2= \left(\dfrac{n + 1}{2}\right)^2$.

For AP with a difference of 2, containing even numbers, the formula is $\left(\dfrac{n}{2}\right) \left(\dfrac{n}{2} + 1\right)$:

$2 + 4 + 6 + … + 2k – 2 + 2k$

$2k + 2k – 2 + … + 6 + 4 + 2$

$\dfrac{(2k + 2 + 2k + 2 + 2k + 2 + … + 2k + 2 + 2k + 2)}{2} = \dfrac{k(2k + 2)}{2}$

$= \dfrac{(2k^2 + 2k)}{2} = 2\dfrac{(k^2 + k)}{2} = k^2 + k =\left( \dfrac{n}{2}\right)^2 + \dfrac{n}{2} = \left(\dfrac{n}{2}\right)\left(\dfrac{n}{2} + 1\right)$. 

Well done to everyone who sent in solutions for this problem!