nth term - what am I supposed to do?

schoolrules1 · 2010-12-03 01:11:39

this is a question from a gcse maths paper.

(a) expand (x+y)^2.

easy, x^2 + 2y + y^2.

(b) Find nth term of 1, 4, 7, 10, 13...

nth term is 3n - 2. also easy...

(c) sophie says that when she squares any term of the arithmetic sequence, she gets an answer which is also a term of the same arithmetic sequence. Use your answers to part (a) and (b) to show that she is correct.

okay, i know that, using (3n-2)^2, i get the sequence: 1, 16, 49, 100, 169, 256... which are all part of the old sequence. but i don't understand how to 'show' or 'prove' that all the numbers in this sequence will always be in the other sequence as well.

help please? thanks

Bob · 2010-12-03 01:34:05

hi schoolrules1

Glad you're ok with the earlier parts because this last part is somewhat more abstract in nature.

Before giving a hint for that look again at

easy, x^2 + 2y + y^2.

There's a little bit missing here. Just a typo I'm sure!

Now for the last part.

(i) Square (3n-2).

(ii) Add then subtract 2 from your expression so you get it in the form (expression) -2

(iii) You should find that the (expression) has 3 as a factor so you can write it as 3 x (another expression) - 2

(iv) The (another expression) is a positive integer (why?) so the term squared is in the same form (3 x [pos. intg] - 2 so it's in the sequence.

Can you fill in the gaps?

Post again if you need more help.

Bob

schoolrules1 · 2010-12-03 01:50:00

hi

thanks for help..the 2y is supposed to be 2xy, sorry

i think i get what you are trying to get me to do. you want me to take (3n-2)^2 and let it be a term in the sequence (3n-2), right?

so i did:

(3n-2)^2 = 9n^2 - 12n + 4

3(9n^2 - 12n + 4) - 2

= 27n^2 - 36n + 12 - 2

= (9n-6)(n-2) - 2

since in the sequence 3n - 2, n is always a positive integer, and 3n - 2 = 4, so (3n - 2) will always be a positive integer.

since it is always a positive whole integer, we can write (9n - 6)(n - 2) - 2 in the form 3x - 2 (where x = (9n - 6)(n - 2)). so any number (3n - 2)^2 is part of the sequence (3n - 2).

is this 'proof' correct? or..is it missing something? i feel like i left something out..

thanks

Bob · 2010-12-03 02:18:45

hi schoolrules1

Right idea. If you can force the expression into the right format you're 99% there.

you wrote:

3(9n^2 - 12n + 4) - 2
= 27n^2 - 36n + 12 - 2
= (9n-6)(n-2) - 2

There's an error in the first line and then you don't need the next two.

3(9n^2 - 12n + 4) - 2 = 27n^2 - 36n +12 -2 which is not what you started with!

Your squaring was right so let's go back to that and manipulate the algebra into the right format.

9n^2 - 12n + 4 = 9n^2 - 12n + 4 + 2 - 2 ..............Here I'm adding 2 then subtracting it again.

That means it's still the original expression.

= (9n^2 -12n +6) - 2 ...............So I've forced out a -2.

There's a factor of 3 that can come out of the first bit.

= 3 x (3n^2 -4n +2) -2 ...............So now it's in the right format for the sequence.

All I need to justify is that 3n^2 - 4n + 2 is always a positive integer.

Well. all it's terms are integers so adding and subtracting them will give another integer.

But how do I know it is positive?

Well, when n = 1 it evaluates to 3 - 4 + 2 = 1 which is positive.

And as n takes values beyond n = 1, 3n^2 gets bigger faster than 4n so it'll still be a positive for n=2, n=3 etc.

So squares of terms in the sequence are also in the form 3M - 2 where M is a positive integer => they're in the sequence too!

Bob

Last edited by Bob (2010-12-03 02:20:45)

Math Is Fun Forum

#1 2010-12-03 01:11:39

nth term - what am I supposed to do?

#2 2010-12-03 01:34:05

Re: nth term - what am I supposed to do?

#3 2010-12-03 01:50:00

Re: nth term - what am I supposed to do?

#4 2010-12-03 02:18:45

Re: nth term - what am I supposed to do?

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