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# help

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If 1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(n*(n + 1)) = 99/100, what is n?

Dec 4, 2019

#1
0

If 1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(n*(n + 1)) = 99/100, what is n?

n = 99, so that:

If 1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(99*(99 + 1)) = 99/100.

Dec 4, 2019
#2
+24364
+1

If

$$\dfrac{1}{1*2} + \dfrac{1}{2*3} + \dfrac{1}{3*4} + \ldots + \dfrac{1}{n*(n + 1)} = \dfrac{99}{100}$$,

what is $$n$$?

$$\begin{array}{|rcll|} \hline \dfrac{99}{100} &=& \dfrac{1}{1*2} + \dfrac{1}{2*3} + \dfrac{1}{3*4} + \ldots + \dfrac{1}{n*(n + 1)} \\ &=& \sum \limits_{m=1}^n \dfrac{1}{m*(m + 1)} &| \quad \dfrac{1}{m*(m + 1)} = \dfrac{1}{m}-\dfrac{1}{m + 1} \\ &=& \sum \limits_{m=1}^n \left( \dfrac{1}{m}-\dfrac{1}{m + 1} \right) \\ &=& \sum \limits_{m=1}^n \dfrac{1}{m} -\sum \limits_{m=1}^n \dfrac{1}{m + 1} \\ &=& \sum \limits_{m=1}^n \dfrac{1}{m} -\sum \limits_{m=2}^{n+1} \dfrac{1}{m} \\ &=& \sum \limits_{m=1}^n \dfrac{1}{m} -\sum \limits_{m=2}^{n} \dfrac{1}{m} - \dfrac{1}{n+1} \\ &=& 1 \underbrace{+ \sum \limits_{m=2}^n \dfrac{1}{m} -\sum \limits_{m=2}^{n} \dfrac{1}{m}}_{=0} - \dfrac{1}{n+1} \\ \mathbf{ \dfrac{99}{100} } &=& \mathbf{ 1 - \dfrac{1}{n+1} } \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{ \dfrac{99}{100} } &=& \mathbf{ 1 - \dfrac{1}{n+1} } \\\\ \dfrac{1}{n+1}&=&1- \dfrac{99}{100} \\\\ \dfrac{1}{n+1}&=& \dfrac{1}{100} \\\\ n+1 &=& 100 \\ n &=& 100-1 \\ \mathbf{n} &=& \mathbf{99} \\ \hline \end{array}$$

Dec 4, 2019