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liberstina [14]
3 years ago
5

Numbers that are 7 units from –3

Mathematics
1 answer:
Sonbull [250]3 years ago
6 0

Answer:

4, and -10

this is because 4 is 7 units more then -3, and -10 id 7 units less then -3

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Solution :

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$           $a_n= \frac{(-1)^n}{n ! 2^n} \ \ \ \ \ a_{n+1}= \frac{(-1)^{n+1}}{(n+1) ! 2^{n+1}}$

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$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$   = $\frac{-1}{1!2}+\frac{1}{2!2^2}-\frac{1}{3! 2^3}+\frac{1}{4! 2^4}-\frac{1}{5! 2^5}$

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