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GREYUIT [131]
3 years ago
9

Approximate the sum of the convergent infinite series the least usingnumber of terms so that the error is less than .0001. What

is the fewest number of terms needed and round your result to three decimal places and state maximum error.
Mathematics
1 answer:
____ [38]3 years ago
3 0

Solution :

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

Error = $|a_{n+1}|$

Error ≤ 0.0001

$|\frac{(-1)^{n+1}}{(n+1)!2^{n+1}}| \leq 0.0001$

$|\frac{1}{(n+1)!2^{n+1}}| \leq 10^{-4}$

$(n+1)! 2^{n+1} \geq 10000$

Now try, n ≥ 5

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$   = $\sum_{n=1 }^{5 } \frac{(-1)^n}{n ! 2^n}$         (with error 0.0001)

 

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

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ = 0.6065104

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a photograph is printed in the center of a piece of paper measuring 7 cm by (x + 8) cm. The photo measures 5 cm by (x +6) cm . T
Nadusha1986 [10]

The paper is 7 cm by (x + 8) cm.

A photo is 5 cm by (x + 6) cm, and is printed in the middle of the paper.

The area of the border/space around the photo is 50 cm².


To find the area of the border around the photo, you find the area of the paper and subtract it by the area of the photo.


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Your equation to find the area of the border around the photo is:

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