C2 math - Little help here?.. Watch

Sokka
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I'm stuck on this geometric progressions question - it's the last one in the book and I've struggled to get here without any help and then i'm baffled by this part, soo, little help plzz?



I did most of part a) here:


They say you have to make the difference between the sum of n terms and the sum to infinity equal to quarter of the nth term. Quarter of the nth term is: 10x0.2^n as shown in my working out. I'm trying to get the last equation shown in my working out EQUAL to 10x0.2^n but I don't know how to manipulate it to get the required result. Basically, how do you get: 10 - 8 x (1-0.2^n / 1-0.2) -----> 10x0.2^n

tyvm if anyone manages to help me!!
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Sokka
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nuodai
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(Original post by Sokka)
They say you have to make the difference between the sum of n terms and the sum to infinity equal to quarter of the nth term. Quarter of the nth term is: 10x0.2^n as shown in my working out. I'm trying to get the last equation shown in my working out EQUAL to 10x0.2^n but I don't know how to manipulate it to get the required result. Basically, how do you get: 10 - 8 x (1-0.2^n / 1-0.2) -----> 10x0.2^n

tyvm if anyone manages to help me!!
You're very close. You can write the denominator of the fraction as 0.8, or \frac{8}{10}, and so using the fact that 1 \div \frac{a}{b} = \frac{b}{a} we get that your expression is equal to 10-8 \times \dfrac{10}{8} \times (1-0.2^n). You can then simplify and you should get the right answer.

(Original post by Sokka)
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Sokka
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i would have never thought of that, thanks
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Sokka
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buump i still don't get how to simplify that part
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Doslargo
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10-8 \times (\dfrac{1-0.2^n}{1-0.2})

Simplifies to

10-8 \times (\dfrac{1-0.2^n}{0.8})

And

\dfrac{8}{0.8} = 10

So it becomes

10-10 \times (1-0.2^n)

Multiply out the brackets to make

10-(10-10 \times 0.2^n)

Then that simplifies to:

10 \times 0.2^n

Which is what you needed, I think.
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