Answer:
The first 6 terms are -7,-14,-28,-56,-112,-224
Step-by-step explanation:
The given sequence is defined recursively by:
and ![a_n=2(a_{n-1})](https://tex.z-dn.net/?f=a_n%3D2%28a_%7Bn-1%7D%29)
When n=2
![a_2=2(a_{2-1})](https://tex.z-dn.net/?f=a_2%3D2%28a_%7B2-1%7D%29)
![a_2=2(a_{1})](https://tex.z-dn.net/?f=a_2%3D2%28a_%7B1%7D%29)
![a_2=2(-7)=-14](https://tex.z-dn.net/?f=a_2%3D2%28-7%29%3D-14)
When n=3
![a_3=2(a_{3-1})](https://tex.z-dn.net/?f=a_3%3D2%28a_%7B3-1%7D%29)
![a_3=2(a_{2})](https://tex.z-dn.net/?f=a_3%3D2%28a_%7B2%7D%29)
![a_3=2(-14)=-28](https://tex.z-dn.net/?f=a_3%3D2%28-14%29%3D-28)
When n=4
![a_4=2(a_{4-1})](https://tex.z-dn.net/?f=a_4%3D2%28a_%7B4-1%7D%29)
![a_4=2(a_{3})](https://tex.z-dn.net/?f=a_4%3D2%28a_%7B3%7D%29)
![a_4=2(-28)=-56](https://tex.z-dn.net/?f=a_4%3D2%28-28%29%3D-56)
When n=5
![a_5=2(a_{5-1})](https://tex.z-dn.net/?f=a_5%3D2%28a_%7B5-1%7D%29)
![a_5=2(a_{4})](https://tex.z-dn.net/?f=a_5%3D2%28a_%7B4%7D%29)
![a_5=2(-56)=-112](https://tex.z-dn.net/?f=a_5%3D2%28-56%29%3D-112)
When n=6
![a_6=2(a_{6-1})](https://tex.z-dn.net/?f=a_6%3D2%28a_%7B6-1%7D%29)
![a_6=2(a_{5})](https://tex.z-dn.net/?f=a_6%3D2%28a_%7B5%7D%29)
![a_6=2(-112)=-224](https://tex.z-dn.net/?f=a_6%3D2%28-112%29%3D-224)
The first 6 terms are -7,-14,-28,-56,-112,-224
<u>Answer:
</u>
An inlet pipe on a swimming pool can be used to fill the pool in 33 hours. Time it will take to fill the pool = 748 hrs
<u>Solution:
</u>
Let x = time it will take to fill the pool
= work rate of inlet pipe
= work rate of drain pipe
Using the formulae
Volume of flow=work rate
time
Let us assume complete time of pool to fill = 1
Remaining fraction of pool to be filled = ![1-\frac{1}{3}=\frac{2}{3}](https://tex.z-dn.net/?f=1-%5Cfrac%7B1%7D%7B3%7D%3D%5Cfrac%7B2%7D%7B3%7D)
![\begin{array}{l}{\frac{1}{33} x-\frac{1}{34} x=\frac{2}{3}} \\\\ {\frac{x}{33}-\frac{x}{34}=\frac{2}{3}}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B%5Cfrac%7B1%7D%7B33%7D%20x-%5Cfrac%7B1%7D%7B34%7D%20x%3D%5Cfrac%7B2%7D%7B3%7D%7D%20%5C%5C%5C%5C%20%7B%5Cfrac%7Bx%7D%7B33%7D-%5Cfrac%7Bx%7D%7B34%7D%3D%5Cfrac%7B2%7D%7B3%7D%7D%5Cend%7Barray%7D)
![\frac{34 x-33 x}{33 \times 34}=\frac{2}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B34%20x-33%20x%7D%7B33%20%5Ctimes%2034%7D%3D%5Cfrac%7B2%7D%7B3%7D)
![34 x-33 x=\frac{2 \times 34 \times 33}{3}](https://tex.z-dn.net/?f=34%20x-33%20x%3D%5Cfrac%7B2%20%5Ctimes%2034%20%5Ctimes%2033%7D%7B3%7D)
![\begin{array}{l}{x=\frac{2244}{3}} \\\\ {x=\frac{3168}{3}} \\\\ {x=748}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7Bx%3D%5Cfrac%7B2244%7D%7B3%7D%7D%20%5C%5C%5C%5C%20%7Bx%3D%5Cfrac%7B3168%7D%7B3%7D%7D%20%5C%5C%5C%5C%20%7Bx%3D748%7D%5Cend%7Barray%7D)
Time it will take to fill the pool = 748 hrs
Answer:
True
Step-by-step explanation:
Hope this helps! :)
With
![y=\displaystyle\sum_{n\ge0}a_nx^{n+r}](https://tex.z-dn.net/?f=y%3D%5Cdisplaystyle%5Csum_%7Bn%5Cge0%7Da_nx%5E%7Bn%2Br%7D)
![y'=\displaystyle\sum_{n\ge0}(n+r)a_nx^{n+r-1}](https://tex.z-dn.net/?f=y%27%3D%5Cdisplaystyle%5Csum_%7Bn%5Cge0%7D%28n%2Br%29a_nx%5E%7Bn%2Br-1%7D)
![y''=\displaystyle\sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-2}](https://tex.z-dn.net/?f=y%27%27%3D%5Cdisplaystyle%5Csum_%7Bn%5Cge0%7D%28n%2Br%29%28n%2Br-1%29a_nx%5E%7Bn%2Br-2%7D)
the singular ODE can be written as
![\displaystyle\sum_{n\ge0}\bigg[2(n+r)(n+r-1)+3(n+r)-1\bigg]a_nx^{n+r}+\sum_{n\ge0}\bigg[-2(n+r)(n+r-1)-3(n+r)\bigg]a_nx^{n+r-1}=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%5Cge0%7D%5Cbigg%5B2%28n%2Br%29%28n%2Br-1%29%2B3%28n%2Br%29-1%5Cbigg%5Da_nx%5E%7Bn%2Br%7D%2B%5Csum_%7Bn%5Cge0%7D%5Cbigg%5B-2%28n%2Br%29%28n%2Br-1%29-3%28n%2Br%29%5Cbigg%5Da_nx%5E%7Bn%2Br-1%7D%3D0)
The first term of the second series admits the indicial equation. When
![n=0](https://tex.z-dn.net/?f=n%3D0)
, we have
![(-2r(r-1)-3r)a_0x^{r-1}=0\iff-2r^2-r=0\iffr^2+\dfrac12r=0](https://tex.z-dn.net/?f=%28-2r%28r-1%29-3r%29a_0x%5E%7Br-1%7D%3D0%5Ciff-2r%5E2-r%3D0%5Ciffr%5E2%2B%5Cdfrac12r%3D0)
Factoring reveals two distinct roots at
![-r(2r+1)=0\implies r_1=0,r_2=-\dfrac12](https://tex.z-dn.net/?f=-r%282r%2B1%29%3D0%5Cimplies%20r_1%3D0%2Cr_2%3D-%5Cdfrac12)
(in your case, swap
![r_1](https://tex.z-dn.net/?f=r_1)
and
![r_2](https://tex.z-dn.net/?f=r_2)
before submitting).
Next, shift the index of the first sum so that it starts at
![n=1](https://tex.z-dn.net/?f=n%3D1)
by replacing
![n\mapsto n-1](https://tex.z-dn.net/?f=n%5Cmapsto%20n-1)
, then consolidate the sums to get
![\displaystyle\sum_{n\ge1}\bigg[2(n+r-1)(n+r-2)+3(n+r-1)-1-2(n+r)(n+r-1)-3(n+r)\bigg]a_nx^{n+r-1}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%5Cge1%7D%5Cbigg%5B2%28n%2Br-1%29%28n%2Br-2%29%2B3%28n%2Br-1%29-1-2%28n%2Br%29%28n%2Br-1%29-3%28n%2Br%29%5Cbigg%5Da_nx%5E%7Bn%2Br-1%7D)
![=\displaystyle x^r\sum_{n\ge1}\bigg[-4n-4r\bigg]a_nx^{n-1}](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%20x%5Er%5Csum_%7Bn%5Cge1%7D%5Cbigg%5B-4n-4r%5Cbigg%5Da_nx%5E%7Bn-1%7D)
Setting
![r=-\dfrac12](https://tex.z-dn.net/?f=r%3D-%5Cdfrac12)
, we then have this as
![=x^{-1/2}\left(-2a_1-6a_2x-10a_3x^2-14a_4x^3-18a_5x^4+\cdots\right)](https://tex.z-dn.net/?f=%3Dx%5E%7B-1%2F2%7D%5Cleft%28-2a_1-6a_2x-10a_3x%5E2-14a_4x%5E3-18a_5x%5E4%2B%5Ccdots%5Cright%29)
![=x^{-1/2}\displaystyle\sum_{n\ge1}(2-4n)a_nx^{n-1}](https://tex.z-dn.net/?f=%3Dx%5E%7B-1%2F2%7D%5Cdisplaystyle%5Csum_%7Bn%5Cge1%7D%282-4n%29a_nx%5E%7Bn-1%7D)
However I don't see the connection to the given answer... It seems some information is missing, specifically about how the coefficients
![a_n](https://tex.z-dn.net/?f=a_n)
are related.
Answer:
The graph in the bottom right corner