The second image is a continuation of the table that might help. Thank you in advance. k2XX- In (1- x) = x+for - 1sx<1kk= 1Σ2+... =33k 2k + 1(-1)*,2* + 1- 1.Xtan'x=x-Σfor |x|<152k + 1k=0d.kP(p - 1)(p - 2)..- (р-k+1)(1 + x)° = Ex*, for |x| <1 and= 1X",k!kk = 0 Identify the function represented by the following power series.001:1+ x +x + .. =1-xE x*, for |x| < 1k= 0k = 0Click the icon to view a table of Taylor series for common functions.1E (- 1)*x*, for |x| <1-x+x + ... =1+Xf(x) =k =0kex = 1+x++... =2!Σfor |x| < 0k!k = 03X(- 1)*x2Σ(2k + 1)!(- 1),2* + 1sin x=x-for |x| < o+<003!5!k= 04k 2k(- 1)*xΣ(2k)!00coS X = 1-for x < o+%3D2!4!k= 02X+23Σk+1 kX'for - 1

Name: The second image is a continuation of the table that might help. Thank
Uploaded: 2024-03-20T06:38:44.000Z
Channel: Bartleby
Description: The second image is a continuation of the table that might help. Thank you in advance. k2XX- In (1- x) = x+for - 1sx<1kk= 1Σ2+... =33k 2k + 1(-1)*,2* + 1- 1.Xtan'x=x-Σfor |x|<152k + 1k=0d.kP(p - 1)(p - 2)..- (р-k+1)(1 + x)° = Ex*, for |x| <1 and= 1X"

Given: The power series ∑k=0∞-1kxk4k It can be written as ∑k=0∞-1kxk4k=∑-1kk=0∞x4k As 11+x=1-x+x2+⋯=∑k=0∞-1kxk Replace x by x4, therefore, ∑k=0∞-1kxk4k=∑-1kk=0∞x4k=11+x4=44+xTherefore, ∑k=0∞-1kxk4k=4x+4.