Pre-calc help

steph7

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Find in terms of a & pi, the sum of the first 50 terms of the series: a+pi, a^2+2pi, a^3+3pi...where a is a fixed real number.
 
What have you tried? Remember, we're here to help you, not to replace you.

I'd split it into two sums: (a + pi) + (a^2 + 2pi) + (a^3 + 3pi) + ... + last term = (a + a^2 + a^3 + ... + last term) + (pi + 2pi + 3pi + ... + last term).

Each part should be a familiar type. Can you find the two sums?
 
Hint: Arrange the terms in the series into a geometric and arithmetic series:

[MATH]S_n=\sum_{k=1}^{n}(a^k)+\pi\sum_{k=1}^{n}(k)[/MATH]
Can you proceed?
 
I already split it apart, I understand how to do the arithmetic part the pi, 2pi and so on. But is the sum for the geometric part the a, a^2... just a? I’m using the formula.
 
I already split it apart, I understand how to do the arithmetic part the pi, 2pi and so on. But is the sum for the geometric part the a, a^2... just a? I’m using the formula.
That "using the formula" business always worries me. There is no substitute for know where it came from and what it actually does.

[math]a + a^2 + a^3 + a^4 + … + a^n = [/math] What? How can you determine that?
 
I already split it apart, I understand how to do the arithmetic part the pi, 2pi and so on. But is the sum for the geometric part the a, a^2... just a? I’m using the formula.
Can you show how you used the formula to get a? Formulas can be good, but you need to think; and it doesn't make a lot of sense that the sum of a lot of positive numbers would equal the first of them, so clearly something is wrong in either the formula or your use of it.
 
I made a mistake so I said a1=a, and r=a. S50= a/(1-a). The formula is a/(1-r). Am I using it wrong? ........edited ...those parentheses () are important.
 
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I made a mistake so I said a1=a, and r=a. S50= a/(1-a). The formula is a/(1-r). Am I using it wrong?
Yes... that equation sum for infinite terms - subject to condition r < 1.

The equation as quoted is incorrect - for limited term series..

Please locate the correct equation for limited terms (n = 50).
 
Yes, with n replaced by 50, of course.

Something you may want to do to convince yourself that you are right is to pick a simple value for a (say 2, or 1/2), and a smaller value for n (say, 5) and calculate the sum both directly and by your formula.
 
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