1. 8 because 8+8=16+1=17
2. 15 because 15-3=12
3. 5 because 5 times 3 =15 and 5times 2=10+5=15
4. 6 because 6times 2 =12 +2=14. And 6+8=14
5. 10 because 6times 10=60
6. 4 because 4+7=11 and 4times2=8+3=11
Answer:
$9.75
Step-by-step explanation:
To get from 4 to a dozen, you have to multiply by 3.
3.25 x 3 = 9.75
3? I think is the answer for this
Answer:
![a(x)=\dfrac{1}{1+2x}](https://tex.z-dn.net/?f=a%28x%29%3D%5Cdfrac%7B1%7D%7B1%2B2x%7D)
Step-by-step explanation:
The generating function a(x) produces a power series ...
![a(x)=a_0+a_1x+a_2x^2+a_3x^3+\dots](https://tex.z-dn.net/?f=a%28x%29%3Da_0%2Ba_1x%2Ba_2x%5E2%2Ba_3x%5E3%2B%5Cdots)
where the coefficients are the elements of the given sequence.
We observe that the given sequence has the recurrence relation ...
![a_0=1;a_n=-2a_{n-1} \quad\text{for n $>$ 0}](https://tex.z-dn.net/?f=a_0%3D1%3Ba_n%3D-2a_%7Bn-1%7D%20%5Cquad%5Ctext%7Bfor%20n%20%24%3E%24%200%7D)
This can be rearranged to ...
![a_n+2a_{n-1}=0](https://tex.z-dn.net/?f=a_n%2B2a_%7Bn-1%7D%3D0)
We can formulate this in terms of a(x) as follows, then solve for a(x).
![\sum\limits^{\infty}_{n=1} {a_{n}x^n} =a(x)-a_0 \quad\text{and}\\\\\sum\limits^{\infty}_{n=1} {2a_{n-1}x^n} =(2x)a(x) \quad\text{so}\\\\\sum\limits^{\infty}_{n=1} {(a_n+2a_{n-1})x^n}=0=a(x)-a_0+2xa(x)\\\\a(x)=\dfrac{a_0}{1+2x}=\dfrac{1}{1+2x}](https://tex.z-dn.net/?f=%5Csum%5Climits%5E%7B%5Cinfty%7D_%7Bn%3D1%7D%20%7Ba_%7Bn%7Dx%5En%7D%20%3Da%28x%29-a_0%20%5Cquad%5Ctext%7Band%7D%5C%5C%5C%5C%5Csum%5Climits%5E%7B%5Cinfty%7D_%7Bn%3D1%7D%20%7B2a_%7Bn-1%7Dx%5En%7D%20%3D%282x%29a%28x%29%20%5Cquad%5Ctext%7Bso%7D%5C%5C%5C%5C%5Csum%5Climits%5E%7B%5Cinfty%7D_%7Bn%3D1%7D%20%7B%28a_n%2B2a_%7Bn-1%7D%29x%5En%7D%3D0%3Da%28x%29-a_0%2B2xa%28x%29%5C%5C%5C%5Ca%28x%29%3D%5Cdfrac%7Ba_0%7D%7B1%2B2x%7D%3D%5Cdfrac%7B1%7D%7B1%2B2x%7D)
The generating function is ...
a(x) = 1/(1+2x)