I need to prove something

[wizzkid67]

prove that the sum of the first 2n multiples of 4 is 4n(2n+1)

[cwes99_03]

Quite simply it is not, as far as I can see.

 

First multiple of 4 would be zero. Since adding zero to anything is pointless we move on to the second 2n multiple which is 8 for n= integers starting with 1 and increasing by 1 for each integer.

 

(2*4)+(4*4)+(6*4)+...+2n*4=?

 

well there are n terms so 4*n would be the median term [(2n*4)/2]. Now the first term and last term are equal to 8+8n, as well as the second and n-1 term and so on and so forth. thus 8+8n=m and m*n/2= the sum of the first 2n multiples if n=even. However if n=odd, then there is an extra term the very middle term that was not paired with another term.

[Queso]

you probably should have named this thread "i need somebody to prove something for me"

because you aren't doing any of the proving.

*sits back down

[arkain101]

and that would make life better..?

[Queso]

what would?

where did "making life better" come into play? :confused:

[goku]

4n(2n+1)

when did they put letters in it :confused:

[Tormod]

when did they put letters in it :confused:

 

It's a nasty invention made by mathematicians and their evil allies in order to confuse those of us who do not want to do our own homework. B)

[sanctus]

let's prove by recurrency (if you don't know this kind of proof just ask) that the sum over the first 2n multiple of 4 is 2*2n(2n+1):

Check for 2n=2 (ie you sum 4 and one "even multiple"): 4*1(2*1+1)=12 OK

Check for 2n=4 (ie you sum 4,8,12 and 16 ):4*2(2*2+1)=40 Ok

 

LEt's suppose it is true for 2(n-1) multiples, let's show that it is also true for 2n: sum over the first 2n multiples of four can be written like the sum over the first 2(n-1)+the two missing terms which can be written as (2n-1)*4 and 2n*4. By hypothesis the sum over the first 2(n-1) terms is 2*2(n-1)(2*(n-1)+1) so the sum over the first 2n terms will be

2*2(n-1)(2*(n-1)+1)+(2n-1)*4+2n=(4n-4)(2n-1)+8n-4+2n=...=2*2n(2n+1) so your sum is proven (as it works for any n).

[wizzkid67]

the first multiple of 4 is not zero its 4 and the second is 8

[Bo]

The proof by induction by sanctus is of course correct (and much more elegent then this proof....), but here follows a formal proof:

you should realize that the factor 4 is irrelevant, also by rescaling n->n/2 the problem simpliefies to showing:

1+2+3+4+5+....n=n/2(n+1)

 

This can be proven in the following way (which is rather ugly, there must be nicer ways...) (S{i=1,n}=sum over i, from 1 to n)

S{i=1,n}(i)=X

1+(n-1)/n+(n-2)/n+...+1/n=X/n

(n+1)-S{i=1,n}(i/n)=X/n

S{i=1,n}(i)=n(n+1)-X

So:

2X=n(n+1) --> X=n/2(n+1)

 

and now do the rest of the homework yorself :confused:

Bo

[sanctus]

the first multiple of 4 is not zero its 4 and the second is 8

 

 

Does it mean you don't like my proof? Or the words used in Cwes's post?

 

Anyway, to prove your thm you are right you have to start by saying that the first multiple is 4.

[sanctus]

This can be proven in the following way (which is rather ugly, there must be nicer ways...)

 

Well, the nicer way would e again induction...:)

[wizzkid67]

lol its not home work at all its just a question i was looking at and yeah i like both the proofs thank u for your time

[snark1100]

1*4+2*4+3*4+...m*4

= 4*(1+2+3+...+m)

= 4*sum from i=0 to m of i

= 4*m(m+1)/2 = 2*m*(m+1)

now, let m=2n to get 4n*(2n+1)