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# Please help, Thank YOU if you try

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Please evaluate$$1/1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+...+2018)$$

### Best Answer

#1
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$$\frac{1}{\frac{1(1+1)}{2}}+\frac{1}{\frac{2(2+1)}{2}}+\dots + \frac{1}{\frac{2018(2018+1)}{2}}$$

$$\frac{2}{1(2)}+\frac{2}{2(3)}+ \dots+\frac{2}{2018(2019)}$$

$$2(\frac{1}{1(2)}+ \frac{1}{2(3)} + \dots + \frac{1}{2018(2019)})$$
$$\text{Apply } \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$

$$2(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3} +\dots -\frac{1}{2019})$$

$$2(1-\frac{1}{2019})$$

$$\frac{4036}{2019}$$

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Apr 4, 2024

### 1+0 Answers

#1
+394
+2
Best Answer

$$\frac{1}{\frac{1(1+1)}{2}}+\frac{1}{\frac{2(2+1)}{2}}+\dots + \frac{1}{\frac{2018(2018+1)}{2}}$$

$$\frac{2}{1(2)}+\frac{2}{2(3)}+ \dots+\frac{2}{2018(2019)}$$

$$2(\frac{1}{1(2)}+ \frac{1}{2(3)} + \dots + \frac{1}{2018(2019)})$$
$$\text{Apply } \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$

$$2(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3} +\dots -\frac{1}{2019})$$

$$2(1-\frac{1}{2019})$$

$$\frac{4036}{2019}$$

hairyberry Apr 4, 2024