STEP I 1992 question 7 solution

Register

About Us | Help | Sign in

Create New Article | Editing Help

Advanced Search

Edit \| Hide
Welcome \| Study Help \| A Levels \| Applications, UCAS etc. \| University Discussion \| Careers \| General Discussion \| Have Your Say \| Health & Relationships \| Debate & Discussion \| Technology \| Sport \| Entertainment \| Chat \| Ask A Moderator

STEP I 1992 question 7 solution

From The Student Room

Jump to:navigation, search

TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 1992 question 7 solution

Solution 1

g(0) = b

and g(1) = a+b

g(n) = an + b

ng(1) = an + bn

g(n) = ng(1) - bn + b = ng(1) - ng(0) + g(0)

g(n) = ng(1) - (n-1)g(0)

As n, g(0) and g(1) are integers, g(n) must be an integer too.

f(-1) = a - b + c

and f(0) = c

and f(1) = a + b + c

and f(n) = an^2 + bn + c

f(-1) + f(1) = 2a + 2c

$\frac{f(-1) + f(1)}{2} - c = a = \frac{f(-1) + f(1)}{2} - f(0)$

So $f(n) = n^2(\frac{f(-1) + f(1)}{2} - f(0)) + bn + c$

$Let\; I = a = \frac{f(-1) + f(1)}{2} - f(0)$

(I is the integer form of a)

nf(1) = an + bn + cn

nf(1) - In - cn = bn

nf(1) - In - nf(0) = bn

We've now got integer forms of an^2, bn and c so f(n) can be expressed as an integer:

f(n) = an^2 + bn + c

f(n) = In^2 + nf(1) - In - nf(0) + c

f(n) = In^2 + nf(1) - In - nf(0) + f(0)

All of which are integers. A more complete f(n):

$f(n) = n^2(\frac{f(-1) + f(1)}{2} - f(0)) + nf(1) - n(\frac{f(-1) + f(1)}{2} - f(0)) - nf(0) + f(0)$

$f(n) = n^2(\frac{f(-1) + f(1)}{2} - f(0)) + nf(1) - n(\frac{f(-1) + f(1)}{2} - f(0)) - f(0)(n - 1)$

And by the by, in case someone wants a justification of why $\frac{f(-1) + f(1)}{2}$ is an integer:

f(-1) + f(1) = 2a + 2c

f(-1) + f(1) = 2(a+c)

The left hand side is an integer. This implies that the right hand side is an integer. As I can take out a factor of 2 out of the right hand side, it implies that the right hand side is an even integer. This then implies that the left hand side is an even integer. And so the left hand side is divisible by 2.

Solution by DeathAwaitsU.

Solution 2

g(x) = ax + b

g(0) = b

g(1) = a+b

g(n) = an+bn

ng(1) = an + bn

g(n) = ng(1) - bn + b

g(n) = ng(1) -n(g(0)) + g(0)

if g(1) and g(0) for any given value of n, g(n) is an integer.

f(x) = Ax^2 + Bx + C

f(-1) = A - B + C

f(0) = C

f(1) = A + B + C

f(n) = An^2 + Bn + C

$f(n) = \frac{1}{2}n^2(f(-1) + f(1) - 2f(0)) + \frac{1}{2}n(f(1)-f(-1)) + f(0)$

Therefore because f(-1),f(0) and f(1) are integers

Now for the bit that harder

$f(\alpha) = A\alpha^2 + B\alpha + C$

$f(\alpha - 1) = A\alpha^2 - 2A\alpha + A + B\alpha - B + C$

$f(\alpha + 1) = A\alpha^2 + 2A\alpha + A + B\alpha + B + C$

$f(\alpha + n) = A\alpha^2 - 2A\alpha n + n^2 + Ba + Bn + C$

Take the first term Aalpha^2

$f(\alpha) - f(\alpha+1) + [f(\alpha) - f(\alpha-1)] = A$

$A = \frac{1}{2}[2f(\alpha) - f(\alpha+1) - f(\alpha-1)]$

$A\alpha^2 = \frac{1}{2}\alpha^2(2f(\alpha) - f(\alpha+1) - f(\alpha-1))$

next term -2A\alpha n

$-2A\alpha n = -\alpha n[2f(\alpha) - f(\alpha+1) - f(\alpha-1)]$

Next two easier terms Ba + C

$B\alpha + C = f(\alpha) - A\alpha^2 = f(\alpha) - \frac{1}{2}\alpha^2[2f(\alpha) - f(\alpha+1) - f(\alpha-1)]$

Next term is Bn

$Bn = \frac{1}{2}n[ (-f(\alpha+1) - f(\alpha-1)) ] - 2\alpha ( 2f(\alpha) - f(\alpha +1) - f(\alpha -1) )]$

So!

$f(\alpha + n) = A\alpha^2 - 2A\alpha n + n^2 + B\alpha + Bn + C$

$= \frac{1}{2}\alpha^2(2f(\alpha) - f(\alpha+1) - f(\alpha-1)) -\alpha n[2f(\alpha) - f(\alpha+1) - f(\alpha-1)] + n^2 + f(\alpha) - \frac{1}{2}\alpha^2[2f(\alpha) - f(\alpha+1) - f(\alpha-1)] + \frac{1}{2}n[ (-f(\alpha+1) - f(\alpha-1)) ] - 2\alpha ( 2f(\alpha) - f(\alpha +1) - f(\alpha -1) )]$

As \alpha and all the functions involving alpha are integers therefore f(\alpha + n) must be an integer.

Solution by insparato.

Retrieved from "http://thestudentroom.co.uk/wiki/STEP_I_1992_question_7_solution"

Article Tools: Create New Article | Report This Article | This Article's History |