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I am trying to find the sum of this -

$$\sum_{r=1}^{100} ( 1 + 2r + 0.3^r ) $$

I know roughly how I am supposed to do. First I distribute the summation across the 3 values.

Then I got stuck $2r$ and $0.3^r$

Both are similar if I understand, so I will only ask for one of them ,

$\sum_{r=1}^{100} (2r)$

how am I suppose to use the property to solve this just like $\sum_{r=1}^{100} = 100(1) = 100 $

Am I right to say -

$\sum_{r=1}^{100} (2r) = 2(1)(100) $ ? But I doubt myself as the r value changes . So I think this is wrong.

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  • $\begingroup$ Try separating the sum into sums of $1$, $2r$, and $0.3^r$ then solving each independently. $\endgroup$ – Shrey Joshi Jul 22 '18 at 17:20
  • $\begingroup$ $\sum_{r=1}^{100} (2r) = (1+2+\cdots+99+100) + (100+99+\cdots+2+1) $ $= (1+100)+(2+99)+\cdots+(99+2)+(100+1) = 100 \times 101$ $\endgroup$ – Henry Jul 22 '18 at 17:25
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You should try breaking up the summation into parts so that you can evaluate them easily

$\displaystyle \sum_{r=1}^{100} ( 1 + 2r + 0.3^r )$

$=\displaystyle \sum_{r=1}^{100}1+\sum_{r=1}^{100}2r+\sum_{r=1}^{100}(0.3)^r$

$\displaystyle = 100+2\sum_{r=1}^{100}r+\sum_{r=1}^{100}(0.3)^r$

$\displaystyle= 100+2\cdot\frac{100(101)}{2}+ \frac{1-(0.3)^{100}}{1-0.3}$

$=\displaystyle 100+10100+1.428 $

$=10201.4285$

EDIT:

You should look up geometric series to understand the summation of $(0,3)^r $

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Hint: $$\sum_{r=1}^{100}{1+2r+0.3^r}=\left(\sum_{r=1}^{100}{1}\right)+\left(\sum_{r=1}^{100}{2r}\right)+\left(\sum_{r=1}^{100}{0.3^r}\right)$$

Solution:

$$\sum_{r=1}^{100}{1+2r+0.3^r}=\left(\sum_{r=1}^{100}{1}\right)+\left(\sum_{r=1}^{100}{2r}\right)+\left(\sum_{r=1}^{100}{0.3^r}\right)$$ $$\left(\sum_{r=1}^{100}{1}\right)=100$$ $$\left(\sum_{r=1}^{100}{2r}\right)=2\sum_{r=1}^{100}{r}=100(101)=10100$$ Since $1+2+3+...+r=r(r+1)/2$ $$\left(\sum_{r=1}^{100}{0.3^r}\right)=\frac{1-0.3^{100}}{1-0.3}$$ The sum is about $10201.428$

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