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Find the sum of the infinite series $$1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots$$

Mar 17, 2020

#1
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The following post is extremely incorrect. If you can identify what is wrong, then great job then help me take down the mathcounts organization.

The formula for infinite series stuff (Mathcounts and AMC 8 has done horrible things to my mind, teaching me to memeroize formulas instead of proving and figuring out myself).

$$\frac{a(1)}{1-r}$$

a(1) = first term

What do you think the first term is? (this one is a brain breaker!)

r = common ratio

Find the common ratio by dividing the second term by the first term.

Plug n' solve!

Mar 17, 2020
edited by CalculatorUser  Mar 19, 2020
#2
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Find the sum of the infinite series

$$1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots$$

$$\small{ \begin{array}{rclllllllllc} & & \mathbf{1}&\mathbf{+} & \mathbf{2\left(\dfrac{1}{1998}\right)} &\mathbf{+} & \mathbf{3\left(\dfrac{1}{1998}\right)^2} &\mathbf{+} &\mathbf{ 4\left(\dfrac{1}{1998}\right)^3} & \mathbf{+}&\mathbf{\cdots} \\\\ & & & & & & & & & & &\mathbf{\text{sum }=\dfrac{a}{1-r}} \\ &=& 1&+ & 1\left(\dfrac{1}{1998}\right)^1 &+ & 1\left(\dfrac{1}{1998}\right)^2 &+ & 1\left(\dfrac{1}{1998}\right)^3 &+&\cdots & \dfrac{1}{1-\dfrac{1}{1998}} \\\\ & & &+ & 1\left(\dfrac{1}{1998}\right)^1 &+ & 1\left(\dfrac{1}{1998}\right)^2 &+ & 1\left(\dfrac{1}{1998}\right)^3 &+&\cdots & \dfrac{\left(\dfrac{1}{1998}\right)^1}{1-\dfrac{1}{1998}} \\\\ & & & & &+ & 1\left(\dfrac{1}{1998}\right)^2 &+ & 1\left(\dfrac{1}{1998}\right)^3 &+&\cdots & \dfrac{\left(\dfrac{1}{1998}\right)^2}{1-\dfrac{1}{1998}} \\\\ & & & & & & & + & 1\left(\dfrac{1}{1998}\right)^3 &+&\cdots & \dfrac{\left(\dfrac{1}{1998}\right)^3}{1-\dfrac{1}{1998}} \\\\ & & & & & & & & & +&\ldots \\ \end{array} }$$

$$\begin{array}{|rcll|} \hline &&\mathbf{ 1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots } \\\\ &=& \dfrac{1}{1-\dfrac{1}{1998}} + \dfrac{\left(\dfrac{1}{1998}\right)^1}{1-\dfrac{1}{1998}} + \dfrac{\left(\dfrac{1}{1998}\right)^2}{1-\dfrac{1}{1998}} + \dfrac{\left(\dfrac{1}{1998}\right)^3}{1-\dfrac{1}{1998}} +\cdots \\\\ &=& \dfrac{1}{1-\dfrac{1}{1998}} \left( 1+ \left(\dfrac{1}{1998}\right)^1+\left(\dfrac{1}{1998}\right)^2+\left(\dfrac{1}{1998}\right)^3 +\cdots \right) \\\\ &=& \dfrac{1}{\dfrac{1998-1}{1998}} \left( 1+ \left(\dfrac{1}{1998}\right)^1+\left(\dfrac{1}{1998}\right)^2+\left(\dfrac{1}{1998}\right)^3 +\cdots \right) \\\\ &=& \dfrac{1998}{1997} \left( 1+ \left(\dfrac{1}{1998}\right)^1+\left(\dfrac{1}{1998}\right)^2+\left(\dfrac{1}{1998}\right)^3 +\cdots \right) \\\\ &=& \dfrac{1998}{1997} \left( \dfrac{1}{1-\dfrac{1}{1998}} \right) \\\\ &=& \dfrac{1998}{1997} \left( \dfrac{1}{\dfrac{1998-1}{1998}} \right) \\\\ &=& \dfrac{1998}{1997}\times \dfrac{1998}{1997} \\\\ &=& \dfrac{1998^2}{1997^2} \\\\ &=&\mathbf{ \dfrac{3992004}{3988009}} \\ \hline \end{array}$$

Mar 17, 2020