Answer:
Radius of convergence of power series is 
Step-by-step explanation:
Given that:
n!! = 1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n n is odd
n!! = 2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n n is even
(-1)!! = 0!! = 1
We have to find the radius of convergence of power series:
![\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\](https://tex.z-dn.net/?f=%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5B%5Cfrac%7B8%5E%7Bn%7Dn%21%283n%2B3%29%21%282n%29%21%21%7D%7B2%5E%7Bn%7D%5B%28n%2B9%29%21%5D%5E%7B3%7D%284n%2B3%29%21%21%7D%5D%288x%2B6%29%5E%7Bn%7D%5C%5C%5C%5C%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5B%5Cfrac%7B8%5E%7Bn%7Dn%21%283n%2B3%29%21%282n%29%21%21%7D%7B2%5E%7Bn%7D%5B%28n%2B9%29%21%5D%5E%7B3%7D%284n%2B3%29%21%21%7D%5D2%5E%7Bn%7D%284x%2B3%29%5E%7Bn%7D%5C%5C%5C%5C%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5B%5Cfrac%7B8%5E%7Bn%7Dn%21%283n%2B3%29%21%282n%29%21%21%7D%7B%5B%28n%2B9%29%21%5D%5E%7B3%7D%284n%2B3%29%21%21%7D%5D%28x%2B%5Cfrac%7B3%7D%7B4%7D%29%5E%7Bn%7D%5C%5C)
Power series centered at x = a is:
![\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\](https://tex.z-dn.net/?f=%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5B%5Cfrac%7B8%5E%7Bn%7Dn%21%283n%2B3%29%21%282n%29%21%21%7D%7B2%5E%7Bn%7D%5B%28n%2B9%29%21%5D%5E%7B3%7D%284n%2B3%29%21%21%7D%5D%288x%2B6%29%5E%7Bn%7D%5C%5C%5C%5C%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5B%5Cfrac%7B8%5E%7Bn%7Dn%21%283n%2B3%29%21%282n%29%21%21%7D%7B2%5E%7Bn%7D%5B%28n%2B9%29%21%5D%5E%7B3%7D%284n%2B3%29%21%21%7D%5D2%5E%7Bn%7D%284x%2B3%29%5E%7Bn%7D%5C%5C%5C%5C%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5B%5Cfrac%7B8%5E%7Bn%7D4%5E%7Bn%7Dn%21%283n%2B3%29%21%282n%29%21%21%7D%7B%5B%28n%2B9%29%21%5D%5E%7B3%7D%284n%2B3%29%21%21%7D%5D%28x%2B%5Cfrac%7B3%7D%7B4%7D%29%5E%7Bn%7D%5C%5C)
![a_{n}=[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}n!(3(n+1)+3)!(2(n+1))!!}{[(n+1+9)!]^{3}(4(n+1)+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]](https://tex.z-dn.net/?f=a_%7Bn%7D%3D%5B%5Cfrac%7B8%5E%7Bn%7D4%5E%7Bn%7Dn%21%283n%2B3%29%21%282n%29%21%21%7D%7B%5B%28n%2B9%29%21%5D%5E%7B3%7D%284n%2B3%29%21%21%7D%5D%5C%5C%5C%5Ca_%7Bn%2B1%7D%3D%5B%5Cfrac%7B8%5E%7Bn%2B1%7D4%5E%7Bn%2B1%7Dn%21%283%28n%2B1%29%2B3%29%21%282%28n%2B1%29%29%21%21%7D%7B%5B%28n%2B1%2B9%29%21%5D%5E%7B3%7D%284%28n%2B1%29%2B3%29%21%21%7D%5D%5C%5C%5C%5Ca_%7Bn%2B1%7D%3D%5B%5Cfrac%7B8%5E%7Bn%2B1%7D4%5E%7Bn%2B1%7D%28n%2B1%29%21%283n%2B6%29%21%282n%2B2%29%21%21%7D%7B%5B%28n%2B10%29%21%5D%5E%7B3%7D%284n%2B7%29%21%21%7D%5D)
Applying the ratio test:
![\frac{a_{n}}{a_{n+1}}=\frac{[\frac{32^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]}{[\frac{32^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]}](https://tex.z-dn.net/?f=%5Cfrac%7Ba_%7Bn%7D%7D%7Ba_%7Bn%2B1%7D%7D%3D%5Cfrac%7B%5B%5Cfrac%7B32%5E%7Bn%7Dn%21%283n%2B3%29%21%282n%29%21%21%7D%7B%5B%28n%2B9%29%21%5D%5E%7B3%7D%284n%2B3%29%21%21%7D%5D%7D%7B%5B%5Cfrac%7B32%5E%7Bn%2B1%7D%28n%2B1%29%21%283n%2B6%29%21%282n%2B2%29%21%21%7D%7B%5B%28n%2B10%29%21%5D%5E%7B3%7D%284n%2B7%29%21%21%7D%5D%7D)
Applying n → ∞
The numerator as well denominator of 
are polynomials of fifth degree with leading coefficients:
(f+g)(x) = f(x) + g(x)
(f+g)(x) = [ f(x) ] + [ g(x) ]
(f+g)(x) = [ 3x-2 ] + [ 2x+1 ]
(f+g)(x) = (3x+2x) + (-2+1)
(f+g)(x) = 5x - 1
Answer is choice B
Answer:
A. Y=2x+3
Step-by-step explanation:
because im right
Answer:
6
Step-by-step explanation:
If p equals 15 then plug 15 in for p and take 15-9=6. pretty sure that's what they are looking for there.
Answer:
...
Step-by-step explanation:
slope= -4/3
y-intercept= -4
y=mx+b
y= -4/3x-4
