Find the sum of the infinite series \(1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots \)
+2864 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!
+26393 Find the sum of the infinite series
\(1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots\)
\(\small{ \begin{array}{rclllllllllc} & & \mathbf{1}&\mathbf{+} & \mathbf{2\left(\dfrac{1}{1998}\right)} &\mathbf{+} & \mathbf{3\left(\dfrac{1}{1998}\right)^2} &\mathbf{+} &\mathbf{ 4\left(\dfrac{1}{1998}\right)^3} & \mathbf{+}&\mathbf{\cdots} \\\\ & & & & & & & & & & &\mathbf{\text{sum }=\dfrac{a}{1-r}} \\ &=& 1&+ & 1\left(\dfrac{1}{1998}\right)^1 &+ & 1\left(\dfrac{1}{1998}\right)^2 &+ & 1\left(\dfrac{1}{1998}\right)^3 &+&\cdots & \dfrac{1}{1-\dfrac{1}{1998}} \\\\ & & &+ & 1\left(\dfrac{1}{1998}\right)^1 &+ & 1\left(\dfrac{1}{1998}\right)^2 &+ & 1\left(\dfrac{1}{1998}\right)^3 &+&\cdots & \dfrac{\left(\dfrac{1}{1998}\right)^1}{1-\dfrac{1}{1998}} \\\\ & & & & &+ & 1\left(\dfrac{1}{1998}\right)^2 &+ & 1\left(\dfrac{1}{1998}\right)^3 &+&\cdots & \dfrac{\left(\dfrac{1}{1998}\right)^2}{1-\dfrac{1}{1998}} \\\\ & & & & & & & + & 1\left(\dfrac{1}{1998}\right)^3 &+&\cdots & \dfrac{\left(\dfrac{1}{1998}\right)^3}{1-\dfrac{1}{1998}} \\\\ & & & & & & & & & +&\ldots \\ \end{array} }\)
\(\begin{array}{|rcll|} \hline &&\mathbf{ 1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots } \\\\ &=& \dfrac{1}{1-\dfrac{1}{1998}} + \dfrac{\left(\dfrac{1}{1998}\right)^1}{1-\dfrac{1}{1998}} + \dfrac{\left(\dfrac{1}{1998}\right)^2}{1-\dfrac{1}{1998}} + \dfrac{\left(\dfrac{1}{1998}\right)^3}{1-\dfrac{1}{1998}} +\cdots \\\\ &=& \dfrac{1}{1-\dfrac{1}{1998}} \left( 1+ \left(\dfrac{1}{1998}\right)^1+\left(\dfrac{1}{1998}\right)^2+\left(\dfrac{1}{1998}\right)^3 +\cdots \right) \\\\ &=& \dfrac{1}{\dfrac{1998-1}{1998}} \left( 1+ \left(\dfrac{1}{1998}\right)^1+\left(\dfrac{1}{1998}\right)^2+\left(\dfrac{1}{1998}\right)^3 +\cdots \right) \\\\ &=& \dfrac{1998}{1997} \left( 1+ \left(\dfrac{1}{1998}\right)^1+\left(\dfrac{1}{1998}\right)^2+\left(\dfrac{1}{1998}\right)^3 +\cdots \right) \\\\ &=& \dfrac{1998}{1997} \left( \dfrac{1}{1-\dfrac{1}{1998}} \right) \\\\ &=& \dfrac{1998}{1997} \left( \dfrac{1}{\dfrac{1998-1}{1998}} \right) \\\\ &=& \dfrac{1998}{1997}\times \dfrac{1998}{1997} \\\\ &=& \dfrac{1998^2}{1997^2} \\\\ &=&\mathbf{ \dfrac{3992004}{3988009}} \\ \hline \end{array}\)
