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Ilya [14]
4 years ago
6

Question 9

Mathematics
1 answer:
Degger [83]4 years ago
7 0
Since x is negative, substitute -x in for x and look for something that would be positive all the time.
A) -x * y = negative
B) x + y = positive, making B the answer
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Find the sum of the first 100 terms in the series
iVinArrow [24]
Hello,

\dfrac{1}{n} - \dfrac{1}{n+1} = \dfrac{1}{n(n+1)} \\

 \dfrac{1}{1*2} = \dfrac{1}{1} - \dfrac{1}{2} \\

 \dfrac{1}{2*3} = \dfrac{1}{2} - \dfrac{1}{3} \\

 \dfrac{1}{3*4} = \dfrac{1}{3} - \frac{1}{4} \\

...\\
 \dfrac{1}{n*(n+1)} = \dfrac{1}{n} - \dfrac{1}{n+1} \\





Adding member by member, we have

\dfrac{1}{1*2} + \dfrac{1}{2*3} +\dfrac{1}{3*4} +...\dfrac{1}{n*(n+1)}=\\
 
\dfrac{1}{1}  - \dfrac{1}{n+1} \\

= \dfrac{n}{n+1} \\



if n=100 sum \boxed{= \dfrac{100}{101} }


3 0
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