Mathematics: (1 point) by recognizing each series below as a taylor series evaluated at a particular value of x, find the sum of each convergent series. a. 1+2+222! +233! +244! +⋯+2nn! +⋯= b. 1−422! +444! −466! +⋯+(−1)n42n(2n)! +⋯=

(1 point) by recognizing each series below as a

Ответ:

tomboyswagge2887

14.03.2022

Guessing on what you're trying to say:

(a)
$1+2+\dfrac{2^2}{2!}+\dfrac{2^3}{3!}+\cdots+\dfrac{2^n}{n!}+\cdots=\displaystyle\sum_{n=0}^\infty\frac{2^n}{n!}$

Compare to the series

$e^x=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}$

which means the value of the sum above is e^2

.

(b)
$1-\dfrac{4^2}{2!}+\dfrac{4^4}{4!}-\cdots+\dfrac{(-1)^n4^{2n}}{(2n)!}+\cdots=\displaystyle\sum_{n=0}^\infty\frac{(-1)^n4^{2n}}{(2n)!}$

Compare to

$\cos x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}$

so that the sum evaluates to $\cos4$ .

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Ответ:

nikejose11

26.03.2023

Mathematics

The area of the rectangle is 120 inches squared.

Step-by-step explanation:

First, identify what you know:

1) The perimeter of the rectangle is 44 inches.

2) The length of the rectangle is 7 more than half its width.

To solve the area of this rectangle, we need to know the width and length of the shape. First, it's important that we know the formula for finding the perimeter, and the area of a rectangle.

Perimeter = 2 x (L + W)

Area = L x W

Now, we can create a formula to demonstrate our problem

Perimeter (44 inches) = 2 (W + ((1/2 x W) + 7)))

Now, we can divide both sides in half, to simplify.

22 inches = (W + ((1/2 x w) + 7))

If you input the values 12, and 10, you'll see that it works!

22 inches = (10 + ((1/2 x 10) + 7))

22 inches = (10 + (5 + 7))

22 inches = (10 + 12)

122 inches = 22 inches

Now, using the Area formula, we can solve for the area of the rectangle, since we know the length and the width.

A = L x W

A = 12 X 10

A = 120 inches squared.

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(1 point) by recognizing each series below as a taylor series evaluated at a particular value of x, find the sum of each convergent series. a. 1+2+222! +233! +244! +⋯+2nn! +⋯= b. 1−422! +444! −466! +⋯+(−1)n42n(2n)! +⋯=

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