Alan

Let f = 1 + 2x + 3x^2 + 4x^5 + ...

First consider y = x + x^2 + x^3 + x^4 + x^5 + ...

y = x(1 + x + x^2 + x^3 + x^4 + ...) or y = x(1 + y)  (as long as y converges, which it does if |x| < 1)

so y(1 - x) = x

y = x/(1 - x)

Now from the original series expression for y we have dy/dx = 1 + 2x + 3x^2 + 4x^3 + ... or in other words

f = dy/dx.  Since y also equals x/(1 - x) we have dy/dx = 1/(1-x) + x/(1-x)^2 or dy/dx = (1-x+x)/(1-x)^2 or dy/dx = 1/(1-x)^2, so, f = 1/(1-x)^2 or

1 + 2x + 3x^2 + 4x^5 + ... = 1/(1 - x)^2 

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In that case the decision is yours to make!

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Yes.  It depends on how many decimal places you have been asked to provide.  5.357% is very slightly more accurate.

To three decimal places it is 5.357%

However, to two decimal places it is 5.36%, because the third digit is greater than 5, so you round up the second digit.

(Don't forget to put the percent sign).

NB: It would have been better to add your response to the same thread as your original question (which could have been done by selecting Post answer there).

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Then the length of 26 of them will be 26 times more than the length of one of them; but since you don't give the length of one of them it's difficult to be more specific!

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percent change is given by 100*(final value - initial value)/initial value, so:

100*(590 - 560)/560

You can crunch the numbers.

What is Colis?

Correct numerical answer Bro, though it would be nice to see your full expressions for x and y!  The reason for this is that there are several ways to solve this problem, each of which can result in expressions that look different, but are, of course, equivalent.

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3x - y = 7             (1)

2x + y = 8            (2)

Add (1) and (2) term by term to eliminate y

5x = 15

Divide both sides by 5

x = 3

Put this back into (2)

2*3 + y = 8   or   6 + y = 8

Subtract 6 from both sides

y = 8 - 6  or  y = 2

You should check these by putting them back into the original equations to see if the left-hand sides equal the right-hand sides.

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$$(5 + \frac{1}{6})\times 7\rightarrow5\times 7 + \frac{1}{6}\times 7\rightarrow 35+\frac{7}{6}\rightarrow35+1+\frac{1}{6}\rightarrow 36\frac{1}{6}$$

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