hey
i've got a few maths problems that i cant seem to solve
Sequences and formulae:
find the value of n for which u has the given value in these:
1. u = 2n u = 256 answer: 8
2. u = 3n u = 2187 answer: 7
3. u = n(n1) u = 380 answer: 20
And then, how to work out the formula for
1. 1/2, 1/3, 1/4
2. 2, 1.5, 1.333333, 1.25
3. 1, 2/3, 3/5, 4/7, 5/9
thanks a lot
Maths Questions (GCSE)
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(Original post by koldtoast)
hey
i've got a few maths problems that i cant seem to solve
Sequences and formulae:
find the value of n for which u has the given value in these:
1. u = 2n u = 256 answer: 8
2. u = 3n u = 2187 7
3. u = n(n1) u = 380 20
And then, how to work out the formula for
1. 1/2, 1/3, 1/4
2. 2, 1.5, 1.333333, 1.25
3. 1, 2/3, 3/5, 4/7, 5/9
thanks a lot 
(Original post by koldtoast)
And then, how to work out the formula for
1. 1/2, 1/3, 1/4
2. 2, 1.5, 1.333333, 1.25
3. 1, 2/3, 3/5, 4/7, 5/9
thanks a lot
for fractions u hav to seperate the top and bottom
1)so 1/2, 1/3, 1/4 would be
1/n1
2)2/1, 3/2, 4/3, 5/4 is the 2nd question in fraction form
so it would be n+1/n
3) 1, 2/3, 3/5, 4/7, 5/9
woulkd be n/the formula for odd numbas...which i cant seem to rememba...or wrk out at the moment!
let me know if ya need explainin on the above 2 
(Original post by Jonatan)
Which types of series are these? Geometric or arithmetic?
the first...iu dunno 
For the first, since its GCSE, you have to use trial and error.
256 = 2^n, try a few numbers and eventually get the right one.
For 3,
380 = n(n1)
380 = n^2  n
0 = n^2  n  380, which you can solve. 
(Original post by koldtoast)
h
And then, how to work out the formula for
1. 1/2, 1/3, 1/4
2. 2, 1.5, 1.333333, 1.25
3. 1, 2/3, 3/5, 4/7, 5/9
thanks a lot
2.) nth term = (n+1)/(n)
3.) nth term = (n)/(2n1) 
(Original post by JamesF)
For the first, since its GCSE, you have to use trial and error.
256 = 2^n, try a few numbers and eventually get the right one.
For 3,
380 = n(n1)
380 = n^2  n
0 = n^2  n  380, which you can solve. 
(Original post by koldtoast)
hey
i've got a few maths problems that i cant seem to solve
Sequences and formulae:
find the value of n for which u has the given value in these:
1. u = 2n u = 256 answer: 8
2. u = 3n u = 2187 answer: 7
3. u = n(n1) u = 380 answer: 20
And then, how to work out the formula for
1. 1/2, 1/3, 1/4
2. 2, 1.5, 1.333333, 1.25
3. 1, 2/3, 3/5, 4/7, 5/9
thanks a lot
1. u = 2n u = 256 answer: 8
the way i see it....
256 = 2n
n = 128
i just dont see how the answer can end up being 8? 
(Original post by koldtoast)
er, i dunno how to solve that lol
Using the quadratic formula:
a = 1, b = 1, c = 380
n = {1(+/)sq.root [(1)^2  (4*1*380)]}/2
n = [1 (+/) sq.root (1 + 1520)]/2
n = [1 + sq. root (1521)]/2 = 20 OR
n = [1  sq.roo t(1521)]/2 = 19
Solutions: n = 20 OR n = 19 
(Original post by koldtoast)
OK thanks i've figured out the second lot of questions involving fractions, but i just can't understand the first question (to do with sequences)
1. u = 2n u = 256 answer: 8
the way i see it....
256 = 2n
n = 128
i just dont see how the answer can end up being 8?
2^8 = 256
You must have copied it down wrong. 
(Original post by koldtoast)
OK thanks i've figured out the second lot of questions involving fractions, but i just can't understand the first question (to do with sequences)
1. u = 2n u = 256 answer: 8
the way i see it....
256 = 2n
n = 128
i just dont see how the answer can end up being 8?
u = 2^n ( 2 to the power n)
Then 8 would be the correct answer.
Btw to solve 3, if you cant solve quadratic equations, then like the others, you will need to use trial and improvement. 
(Original post by JamesF)
I thought you had just misread the question, if it said
u = 2^n ( 2 to the power n)
Then 8 would be the correct answer.
Btw to solve 3, if you cant solve quadratic equations, then like the others, you will need to use trial and improvement. 
(Original post by koldtoast)
OK thanks i've figured out the second lot of questions involving fractions, but i just can't understand the first question (to do with sequences)
1. u = 2n u = 256 answer: 8
the way i see it....
256 = 2n
n = 128
i just dont see how the answer can end up being 8? 
(Original post by Undry1)
i think the question is supposed to be 2^n rather than 2n. 
(Original post by bono)
n^2  n  380 = 0
Using the quadratic formula:
a = 1, b = 1, c = 380
n = {1(+/)sq.root [(1)^2  (4*1*380)]}/2
n = [1 (+/) sq.root (1 + 1520)]/2
n = [1 + sq. root (1521)]/2 = 20 OR
n = [1  sq.roo t(1521)]/2 = 19
Solutions: n = 20 OR n = 19 
hmm, i'm typing it straight out of the book, which btw is the Edexcel GCSE Maths Higher Course book for 2001 specs... page 292 exercise 14b
Q8. u = 2n u= 256
Q9. u = 3n u= 2187
Q10.u = n(n1) u= 380 
(Original post by koldtoast)
hmm, i'm typing it straight out of the book, which btw is the Edexcel GCSE Maths Higher Course book for 2001 specs... page 292 exercise 14b
Q8. u = 2n u= 256
Q9. u = 3n u= 2187
Q10.u = n(n1) u= 380
Can't you see it's "2^" ? 
2^n makes sense, and that is what the answer at the back would seem to support.
its weird but it REALLY looks like 2n instead of 2^n (with a little n towards the top right of the '2')... i'll see if i can scan the page somehow or something 
(Original post by koldtoast)
2^n makes sense, and that is what the answer at the back would seem to support.
its weird but it REALLY looks like 2n instead of 2^n (with a little n towards the top right of the '2')... i'll see if i can scan the page somehow or something
If you can, scan in the page, i'd be interested to see it. 
I just checked in my book and he/she is right the question is as they typed it...