Please help with c1 sequences and series porblem

show that the sum of the first 2n natural numbers is n(2n+1)

Ive seen the example in the c1 book.
where it talks of the first n numbers sum is 1/2n(n+1)
I can see the process but cant see the logic to find the first to as that would be 3 and when calculated I get..
s=n/2[2a+(n-1)d]
s=2/2[2x1+(2-1)1]
s=1[2+1]
this is figuratively correct
but obviously not right
please help..

Reply 1

Overmars

Sorry, I'm not exactly sure what you're confused about. You can do the question, but you don't understand why it works? If so, what bit don't you understand?

Reply 2

Dadeyemi

what do you mean?
they are using the sum of an arithmetic sequence equation what might be confusing is that they have set n = 2n let for sake of simplicity turn this to n = 2k and as the first 2n numbers starts with 1 a = 1 and it is all the integers 1, 2 , 3... 2n so the common difference, d = 1

Substitute these into:

S=(n/2)*(2a+(n-1)d)

=>
s=(2k/2)*(2n*1+(2-1)1)
s=k(2k+1)
replace k = n
Sum_{1,2,3,..,2n - 1,2n} = n(2n+1)
as required

hope that helps you understand

Reply 3

username21537

Argue this by induction - Show that it holds for n=1. then assuming it holds for some n, that it holds for n+1. This will involve adding the n+1'st term and showing that it ends up in the same form but with n's replaced by n+1's, n-1's replaced with n's and so on.

Reply 4

TSR George

Many thanks to all your help..

My problem understanding was that the question asks for the first 2n, which I took it as asking for the first 2 numbers 1+2 whose sum I believed was 3.

when calculated It I wrote
s=n/2[2a+(n-1)d]
s=2/2[2x1+(2-1)1]
s=1[2+1] ....what confused me was I lost the n factor..

As the first two n numbers are 1 and 2. The sum of which should be 3. I only used 2 as it got the right shape for the answer, after calculating the sum equaled s=1[2+1] I lost the n definition.

With all your help I think I now understand, 2n was just asking for the first 2 numbers but with n as the factor of 2...
so if I understand right what I should have written was
s=2n/2[2a+(2n-1)d]
s=2n/2[2x1+(2n-1)1]
s=n[2+2n-1]
s=n[2n+1]

It looks right, and I understand it, thank you to you all for your wonderful help, this site is the best.
Elli

Reply 5

Leonb

It will be confusing if you interpret the first 2n natural numbers as the set of even numbers i.e n=1 2n=2, n=2 2n=4 (like I did!), when in fact all it is asking for is the first x of the set of natural numbers. The natural numbers being 1,2,3,4... , edexcel have just chosen to confuse things by using the term 2n instead of x.
Simply use the formula for finding the sum of a series: n/2(2a+(n-1)d).
For this question a=1, d=1, n=2n.
So: 2n/2(2+2n-1)
= (4n+4n^2-2n)/2
= (2n+4n^2)/2
= (n+2n^2)
= n(1 +2n) OR n(2n+1)