HL Maths, 1994, Question 2, Part C

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If, for all integers n,

Un = 600(2)^n - 7(5)^n

Verify that

Un+2 - 7Un+1 + 10Un = 0

Find the least integer n > 0 for which Un < 0

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I've gotten the first part perfectly, but the "Find the least..." part is annoying me I can't seem to get anywhere with it! Bah! Algebra! My papers say the answer is 22 but I'm getting nowhere. Can anyone type up a solution to this?

My working so far has been:

Un < 0
600(2)^n - 7(5)^n < 0
600(2)^n < 7(5)^n
600 < 7(5/2)^n

But I don't think I'm getting anywhere.

*Angel*

Are you sure that's '94?

Anway you're given Un = 600(2^n) - 7(5^n)

you need to find U(n+1) and U(n+2)

U(n+1) = 600(2)(2^n) - 7(5)(5^n)
= 1200(2^n) - 35(5^n)

do the same for U(n+2), then sub the two in where appropriate to show the equation given is equal to zero.

Sev

Ok.. we'll ive done this an entirely different way,

if A > B
then logA > logB

so..

if (600/7) < (2.5)^n

as you have already worked out

then take the log (base 2.5) of both sides

log(2.5)[600/7] < log(2.5)[(5/2)^n]

log(2.5)[600/7] < n

Now to work out the base 2.5 logarithm of 600/7 on a calculator, you can use the identity that

log(a)[x] = log(b)[x]/log(b)[a]

where b is an arbitrary number

so for simplicity, let b = e = 2.73, then we can just use the natural logarithm, 'ln', on the calculator

so log(2.5)[600/7] = ln[600/7]/ln[2.5]

= 4.858

so n > 4.858

but since they asked you for the least integer, then the answer is 5.

*Angel*

Oops I didn't read the question fully, btw it's the '94 sample paper.

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Ah, it is the sample paper.... christ that was annoying me today... I had tried it using logs and the change of base formula but because I *thought* I was doing 1994's paper I thought the answer was... ah nevermind. Take study breaks kids! Thanks for your help guys.