HL Maths- Sequences and Series problem

DmanDmythDledge

Anyone able to explain the answer of this to me? I'm posting solution but don't know how it was gotten.

Express in terms of n:
(i) 1 + 2 + 3 + ... + 2n
(ii) 2+ 4 + 6 + ... + 2n
(iii) 1 + 3 + 5 + ... + (2n-1).

Solutions:

(i) 2n(2n+1)/2

=2n^2 + n.

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

=2(n)(n+1)/2
=n^2 + n

(iii) (1 + 2 + ... + 2n) - (2 + 4 + ... +2n)

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

They seem like some sort of arithmetic series but using the formula of Sn doesn't get the answer.

yummy91

what Sn formula are you using cause they worked out for me.
I used sn= n/2(2a+(n-1)d)

(i) d=1
a=1
n=2n
Sn= 2n/2 (2(1)+(2n-1)1)
= 2n(2+2n-1)/2
= 2n(2n+1)/2

Pygmalion

They are arithmetic. Just looking at them I'd say:

1) a = 1, d = 1, but it goes to 2n, so double what you get for Sn.

2) a = 2, d = 2, it goes to 2n, but since it's going up in 2s it only has n terms, so the Sn formula should work.

3) a = 1, d = 2, goes to (2n - 1), but again, that's just n terms, I'd imagine Sn formula should work.

I haven't done these out though so might have made stupid mistake (might go get my copy and try them)

SligoBrewer

ll do them and scan them for you dman

ironclaw

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

a = 1 , d =1

(ii) 2+ 4 + 6 + ... + 2n

a = 2, d = 1

(iii) 1 + 3 + 5 + ... + (2n-1).

a=1 d=2

Sub them into: (n/2)(2a+(n-1)d)

EDIT: Was beaten to it!

SligoBrewer

1 worked out.:D

The others however....:o

Dubs

theyre the natural number ones

formulas are:
En = 1 + 2 + 3 ... + n = (n/2)(n + 1)

and

En^2 = (1^2) + (2^2) + (3^2) ... (n^2) = (n/6)(n + 1)(2n + 1)

SligoBrewer

Dubs wrote: »

theyre the natural number ones

formulas are:
En = 1 + 2 + 3 ... + n = (n/2)(n + 1)

and

En^2 = (1^2) + (2^2) + (3^2) ... (n^2) = (n/6)(n + 1)(2n + 1)

There we are!

Well spotted.:D

What do you do for 3 though?

The first one works out all natural numbers.

The second one works out all even numbers.

The third part asks you to find the sum of the odd numbers.

So, sum of odd = (sum of all) - (sum of even).

yummy91

Dubs wrote: »

theyre the natural number ones

formulas are:
En = 1 + 2 + 3 ... + n = (n/2)(n + 1)

damn it i got it wrong!!
really really hope that that doesnt happen tomorrow.... hehe.... sooo f***ed:o:P

DmanDmythDledge

Nice one lads.

Now to go back over everything in differentiation and integration and I'm sorted. Going to be a long night.:(

Killaqueen!!!

Sequences and series are so hard it's annoying that it's two questions on the paper - I have no choice now. Still gave it a go today thought and am trying to learn the like 10 formulas for sequence and series

Why don't they give them on the tables!!!

Killaqueen!!!

Killaqueen!!! wrote: »

Sequences and series are so hard it's annoying that it's two questions on the paper - I have no choice now. Still gave it a go today thought and am trying to learn the like 10 formulas for sequence and series

Why don't they give them on the tables!!!

It's only one question: Question 4. Although, very rarely, bits of it crop up in 1/2/5 - very rarely, though.

SligoBrewer

Killaqueen!!! wrote: »

Sequences and series are so hard it's annoying that it's two questions on the paper - I have no choice now. Still gave it a go today thought and am trying to learn the like 10 formulas for sequence and series

Why don't they give them on the tables!!!

Well its not really two full questions.
Especially when you factor [(pun) :pac:] in Logs, Binomial and Induction.

Killaqueen!!!

Yeah, but I suck at Q5 anyway because it not only has sequences and series but, as you mentioned, binomial, logs etc. which I also suck at

I'm actually writing out a list of formulaes needed for those two questions and it's unbelievable. I know you're not supposed to have to remember something like the binomial (it's like the -b formula you're supposed to know it from doing it) but....

Un= Sn - Sn-1

Arithmetic:

Un=a + (n-1)d

Sn= n/2 [2a+(n-1)d]

Test for Arithmetic: Un+1 - Un = Constant

Geometric

Un= ar^n-1

Sn= a(1-r^n) / 1-r

Sum to infinity= a/1-r

Test for Geometric: Un+1/Un = Constant


Series

Sum r = n/2 (n+1)

Sum r squared: n/6 (n+1)(2n+1)


Binomial

The binomial theorem, its general term and middle term



Like seriously, too much

TheManWho

I personally think it's the best topic on the paper, and I can say all of those formulas from memory with ease. But....hmm....that doesn't help you much does it

Zonda999

If anyone is looking over that question, i advise looking over the telescopic series, and the arithmetico-geometric series. The latter may be a part C but theyre actually rather easy. Hopin for a good Q4 & 5 tomorrow. Then the plan is to leave out one of the algebra(Probably the 2nd one judging by past papers) and possibly one of the differentiation questions