Answer:
1/13
Step-by-step explanation:
Since 78 is divisible by 6, then 6/78 reduces to 1/13.
(y-2)4 + 21
You subtract two because the 21 accounts for it and then whatever you have left multiplied by 4 and added to 21 is the amount of years after conversion.
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Answer:
![\sum^\infty_{n=0} -5 (\frac{x+2}{2})^n](https://tex.z-dn.net/?f=%5Csum%5E%5Cinfty_%7Bn%3D0%7D%20-5%20%28%5Cfrac%7Bx%2B2%7D%7B2%7D%29%5En)
Step-by-step explanation:
Rn(x) →0
f(x) = 10/x
a = -2
Taylor series for the function <em>f </em>at the number a is:
![f(x) = \sum^\infty_{n=0} \frac{f^{(n)}(a)}{n!} (x - a)^n](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20%5Csum%5E%5Cinfty_%7Bn%3D0%7D%20%5Cfrac%7Bf%5E%7B%28n%29%7D%28a%29%7D%7Bn%21%7D%20%28x%20-%20a%29%5En)
............ equation (1)
Now we will find the function <em>f </em> and all derivatives of the function <em>f</em> at a = -2
f(x) = 10/x f(-2) = 10/-2
f'(x) = -10/x² f'(-2) = -10/(-2)²
f"(x) = -10.2/x³ f"(-2) = -10.2/(-2)³
f"'(x) = -10.2.3/x⁴ f'"(-2) = -10.2.3/(-2)⁴
f""(x) = -10.2.3.4/x⁵ f""(-2) = -10.2.3.4/(-2)⁵
∴ The Taylor series for the function <em>f</em> at a = -4 means that we substitute the value of each function into equation (1)
So, we get
Or ![\sum^\infty_{n=0} -5 (\frac{x+2}{2})^n](https://tex.z-dn.net/?f=%5Csum%5E%5Cinfty_%7Bn%3D0%7D%20-5%20%28%5Cfrac%7Bx%2B2%7D%7B2%7D%29%5En)