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

17

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

13

ellishnoo

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..

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

2

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

ellishnoo

OP

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)

Please help with c1 sequences and series porblem

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