Respuesta :
Guessing on what you're trying to say:
(a)
[tex]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!}[/tex]
Compare to the series
[tex]e^x=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}[/tex]
which means the value of the sum above is [tex]e^2[/tex].
(b)
[tex]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)!}[/tex]
Compare to
[tex]\cos x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}[/tex]
so that the sum evaluates to [tex]\cos4[/tex].
(a)
[tex]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!}[/tex]
Compare to the series
[tex]e^x=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}[/tex]
which means the value of the sum above is [tex]e^2[/tex].
(b)
[tex]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)!}[/tex]
Compare to
[tex]\cos x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}[/tex]
so that the sum evaluates to [tex]\cos4[/tex].
The sum of (a) series is [tex]\rm e^2[/tex] and for (b) the sum is cos4.
It is given that the taylor series evaluated at a particular value of x.
It is required to find the sum of the convergent series given.
What is taylor series?
It is defined as the sum of the infinite derivative of functions at a single point the taylor series will give an approximate value of a function around a certain point.
We have series:
[tex]=\rm 1+2+\frac{2^2}{2!} +\frac{3^2}{3!} +\frac{4^2}{4!} +......[/tex]
It can be written as in the summation form:
[tex]=\rm \sum_{n=0}^{\infty}\frac{2^n}{n!}[/tex]
We know the taylor expansion of the [tex]\m e^x[/tex]:
[tex]\rm e^x=\rm \sum_{n=0}^{\infty}\frac{2^n}{n!}[/tex]
Compare both series, we get:
The sum of the given series:
[tex]\rm e^2 =\rm 1+2+\frac{2^2}{2!} +\frac{3^2}{3!} +\frac{4^2}{4!} +......[/tex]
For the series:
[tex]1-\frac{4^2}{2!} +\frac{4^4}{4!} -....\frac{(-1)^n4^2^n}{(2n)!} +....[/tex]
It can be written as in the summation form:
[tex]\rm e^x=\rm \sum_{n=0}^{\infty}\frac{(-1)^n4^2^n}{2n!}[/tex]
If we compare this with taylor expansion of cosx, we get:
[tex]\rm cos4 =1-\frac{4^2}{2!} +\frac{4^4}{4!} -.....[/tex]
Thus, the sum of (a) series is [tex]\rm e^2[/tex] and for (b) the sum is cos4.
Learn more about the taylor series here:
https://brainly.com/question/26549116
