Mathematics, 05.07.2021 19:40 Maddifalls908

Find an expression for the general term of each of the series below. Use n as your index, and pick your general term so that the sum giving the series starts with n=0. A. x^3cosx^2=x^3-(x^7)/2!+(x^11)/4!-(x ^15)/6!+...
general term =
B. x^3sinx^2=x^5-(x^9)/3!+(x^13)/5!-(x ^17)/7!+...
general term =

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Answer from: lashondrascott

$x^{3}cos(x^{2})=\sum _{n=0} ^{\infty} \frac{(-1)^{n}x^{4n+3}}{(2n)!}$

$x^{3}sin(x^{2})=\sum _{n=0} ^{\infty} \frac{(-1)^{n}x^{4n+5}}{(2n+1)!}$

Step-by-step explanation:

Let's start with the first function:

$x^{3}cos(x^{2})=x^{3}-\frac{x^{7}}{2!}+\frac{x^{11}}{4!}-\frac{x^{15}}{6!}+...$

In order to find the expression for the general term, we will need to analyze each part of the sum. First, notice that the sign of the terms of the sum will change with every new term, this tells us that the expression must contain a

$(-1)^{n}$ .

This will guarantee us that the terms will always change their signs so that will be the first part of our expression.

next, the power of the x. Notice the given sequence: 3, 7, 11, 15...

we can see this is an arithmetic sequence since the distance between each term is the same. There is a distance of 4 between each consecutive power, so this sequence can be found by adding a 4n to the original number, the 3. So the power is given by 4n+3.

so let's put the two things together:

$(-1)^{n}x^{4n+3}$

Finally the denominator, there is also a sequence there: 0!, 2!, 4!, 6!

This is also an arithmetic sequence, where we are multiplying each consecutive value of n by a 2, so in this case the sequence can be written as: (2n)!

So let's put it all together so we get:

$\frac{(-1)^{n}x^{4n+3}}{(2n)!}$

So now we can build the whole series:

$x^{3}cos(x^{2})=\sum _{n=0} ^{\infty} \frac{(-1)^{n}x^{4n+3}}{(2n)!}$

Now, let's continue with the next function:

$x^{3}sin(x^{2})=x^{5}-\frac{x^{9}}{3!}+\frac{x^{13}}{5!}-\frac{x^{17}}{7!}+...$

$(-1)^{n}$ .

This will guarantee us that the terms will always change their signs so that will be the first part of our expression.

next, the power of the x. Notice the given sequence: 5, 9, 13, 17...

so let's put the two things together:

$(-1)^{n}x^{4n+5}$

Finally the denominator, there is also a sequence there: 1!, 3!, 5!, 7!

This is also an arithmetic sequence, where we are multiplying each consecutive value of n by a 2 starting from a 1, so in this case the sequence can be written as: (2n+1)!

So let's put it all together so we get:

$\frac{(-1)^{n}x^{4n+5}}{(2n+1)!}$

So now we can build the whole series:

$x^{3}sin(x^{2})=\sum _{n=0} ^{\infty} \frac{(-1)^{n}x^{4n+5}}{(2n+1)!}$

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