Answer:
300
Step-by-step explanation:
Answer: 28
Step-by-step explanation: i got the same test
![\displaystyle y=\sum_{n\ge0}a_nx^n](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3D%5Csum_%7Bn%5Cge0%7Da_nx%5En)
![\implies\displaystyle y''=\sum_{n\ge2}n(n-1)a_nx^{n-2}](https://tex.z-dn.net/?f=%5Cimplies%5Cdisplaystyle%20y%27%27%3D%5Csum_%7Bn%5Cge2%7Dn%28n-1%29a_nx%5E%7Bn-2%7D)
![y''-3xy=0](https://tex.z-dn.net/?f=y%27%27-3xy%3D0)
![\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-3\sum_{n\ge0}a_nx^{n+1}=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%5Cge2%7Dn%28n-1%29a_nx%5E%7Bn-2%7D-3%5Csum_%7Bn%5Cge0%7Da_nx%5E%7Bn%2B1%7D%3D0)
![\displaystyle2a_2+\sum_{n\ge3}n(n-1)a_nx^{n-2}-3\sum_{n\ge0}a_nx^{n+1}=0](https://tex.z-dn.net/?f=%5Cdisplaystyle2a_2%2B%5Csum_%7Bn%5Cge3%7Dn%28n-1%29a_nx%5E%7Bn-2%7D-3%5Csum_%7Bn%5Cge0%7Da_nx%5E%7Bn%2B1%7D%3D0)
![\displaystyle2a_2+\sum_{n\ge3}n(n-1)a_nx^{n-2}-3\sum_{n\ge3}a_{n-3}x^{n-2}=0](https://tex.z-dn.net/?f=%5Cdisplaystyle2a_2%2B%5Csum_%7Bn%5Cge3%7Dn%28n-1%29a_nx%5E%7Bn-2%7D-3%5Csum_%7Bn%5Cge3%7Da_%7Bn-3%7Dx%5E%7Bn-2%7D%3D0)
![\displaystyle2a_2+\sum_{n\ge3}\bigg(n(n-1)a_n-3a_{n-3}\bigg)x^{n-2}=0](https://tex.z-dn.net/?f=%5Cdisplaystyle2a_2%2B%5Csum_%7Bn%5Cge3%7D%5Cbigg%28n%28n-1%29a_n-3a_%7Bn-3%7D%5Cbigg%29x%5E%7Bn-2%7D%3D0)
This generates the recurrence relation
![\begin{cases}a_0=a_0\\a_1=a_1\\2a_2=0\\n(n-1)a_n-3a_{n-3}=0&\text{for }n\ge3\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Da_0%3Da_0%5C%5Ca_1%3Da_1%5C%5C2a_2%3D0%5C%5Cn%28n-1%29a_n-3a_%7Bn-3%7D%3D0%26%5Ctext%7Bfor%20%7Dn%5Cge3%5Cend%7Bcases%7D)
Because you have
![n(n-1)a_n-3a_{n-3}=0\implies a_n=\dfrac3{n(n-1)}a_{n-3}](https://tex.z-dn.net/?f=n%28n-1%29a_n-3a_%7Bn-3%7D%3D0%5Cimplies%20a_n%3D%5Cdfrac3%7Bn%28n-1%29%7Da_%7Bn-3%7D)
it follows that
![a_2=0\implies a_5=a_8=a_{11}=\cdots=a_{n=3k-1}=0](https://tex.z-dn.net/?f=a_2%3D0%5Cimplies%20a_5%3Da_8%3Da_%7B11%7D%3D%5Ccdots%3Da_%7Bn%3D3k-1%7D%3D0)
for all
![k\ge1](https://tex.z-dn.net/?f=k%5Cge1)
.
For
![n=1,4,7,10,\ldots](https://tex.z-dn.net/?f=n%3D1%2C4%2C7%2C10%2C%5Cldots)
, you have
![a_1=a_1](https://tex.z-dn.net/?f=a_1%3Da_1)
![a_4=\dfrac3{4\times3}a_1=\dfrac{3\times2}{4!}a_1](https://tex.z-dn.net/?f=a_4%3D%5Cdfrac3%7B4%5Ctimes3%7Da_1%3D%5Cdfrac%7B3%5Ctimes2%7D%7B4%21%7Da_1)
![a_7=\dfrac3{7\times6}a_4=\dfrac{3\times5}{7\times6\times5}=\dfrac{3^2\times5\times2}{7!}](https://tex.z-dn.net/?f=a_7%3D%5Cdfrac3%7B7%5Ctimes6%7Da_4%3D%5Cdfrac%7B3%5Ctimes5%7D%7B7%5Ctimes6%5Ctimes5%7D%3D%5Cdfrac%7B3%5E2%5Ctimes5%5Ctimes2%7D%7B7%21%7D)
![a_{10}=\dfrac3{10\times9}a_7=\dfrac{3\times8}{10\times9\times8}a_7=\dfrac{3^3\times8\times5\times2}{10!}a_1](https://tex.z-dn.net/?f=a_%7B10%7D%3D%5Cdfrac3%7B10%5Ctimes9%7Da_7%3D%5Cdfrac%7B3%5Ctimes8%7D%7B10%5Ctimes9%5Ctimes8%7Da_7%3D%5Cdfrac%7B3%5E3%5Ctimes8%5Ctimes5%5Ctimes2%7D%7B10%21%7Da_1)
so that, in general, for
![n=3k-2](https://tex.z-dn.net/?f=n%3D3k-2)
,
![k\ge1](https://tex.z-dn.net/?f=k%5Cge1)
, you have
![a_{n=3k-2}=\dfrac{3^{k-1}\displaystyle\prod_{\ell=1}^{k-1}(3\ell-1)}{(3k-2)!}a_1](https://tex.z-dn.net/?f=a_%7Bn%3D3k-2%7D%3D%5Cdfrac%7B3%5E%7Bk-1%7D%5Cdisplaystyle%5Cprod_%7B%5Cell%3D1%7D%5E%7Bk-1%7D%283%5Cell-1%29%7D%7B%283k-2%29%21%7Da_1)
Now, for
![n=0,3,6,9,\ldots](https://tex.z-dn.net/?f=n%3D0%2C3%2C6%2C9%2C%5Cldots)
, you have
![a_0=a_0](https://tex.z-dn.net/?f=a_0%3Da_0)
![a_3=\dfrac3{3\times2}a_0=\dfrac3{3!}a_0](https://tex.z-dn.net/?f=a_3%3D%5Cdfrac3%7B3%5Ctimes2%7Da_0%3D%5Cdfrac3%7B3%21%7Da_0)
![a_6=\dfrac3{6\times5}a_3=\dfrac{3\times4}{6\times5\times4}a_3=\dfrac{3^2\times4}{6!}a_0](https://tex.z-dn.net/?f=a_6%3D%5Cdfrac3%7B6%5Ctimes5%7Da_3%3D%5Cdfrac%7B3%5Ctimes4%7D%7B6%5Ctimes5%5Ctimes4%7Da_3%3D%5Cdfrac%7B3%5E2%5Ctimes4%7D%7B6%21%7Da_0)
![a_9=\dfrac3{9\times8}a_6=\dfrac{3\times7}{9\times8\times7}a_6=\dfrac{3^3\times7\times4}{9!}a_0](https://tex.z-dn.net/?f=a_9%3D%5Cdfrac3%7B9%5Ctimes8%7Da_6%3D%5Cdfrac%7B3%5Ctimes7%7D%7B9%5Ctimes8%5Ctimes7%7Da_6%3D%5Cdfrac%7B3%5E3%5Ctimes7%5Ctimes4%7D%7B9%21%7Da_0)
and so on, with a general pattern for
![n=3k](https://tex.z-dn.net/?f=n%3D3k)
,
![k\ge0](https://tex.z-dn.net/?f=k%5Cge0)
, of
![a_{n=3k}=\dfrac{3^k\displaystyle\prod_{\ell=1}^k(3\ell-2)}{(3k)!}a_0](https://tex.z-dn.net/?f=a_%7Bn%3D3k%7D%3D%5Cdfrac%7B3%5Ek%5Cdisplaystyle%5Cprod_%7B%5Cell%3D1%7D%5Ek%283%5Cell-2%29%7D%7B%283k%29%21%7Da_0)
Putting everything together, we arrive at the solution
![y=\displaystyle\sum_{n\ge0}a_nx^n](https://tex.z-dn.net/?f=y%3D%5Cdisplaystyle%5Csum_%7Bn%5Cge0%7Da_nx%5En)
![y=a_0\underbrace{\displaystyle\sum_{k\ge0}\frac{3^k\displaystyle\prod_{\ell=1}^k(3\ell-2)}{(3k)!}x^{3k}}_{n=0,3,6,9,\ldots}+a_1\underbrace{\displaystyle\sum_{k\ge1}\frac{3^{k-1}\displaystyle\prod_{\ell=1}^{k-1}(3\ell-1)}{(3k-2)!}x^{3k-2}}_{n=1,4,7,10,\ldots}](https://tex.z-dn.net/?f=y%3Da_0%5Cunderbrace%7B%5Cdisplaystyle%5Csum_%7Bk%5Cge0%7D%5Cfrac%7B3%5Ek%5Cdisplaystyle%5Cprod_%7B%5Cell%3D1%7D%5Ek%283%5Cell-2%29%7D%7B%283k%29%21%7Dx%5E%7B3k%7D%7D_%7Bn%3D0%2C3%2C6%2C9%2C%5Cldots%7D%2Ba_1%5Cunderbrace%7B%5Cdisplaystyle%5Csum_%7Bk%5Cge1%7D%5Cfrac%7B3%5E%7Bk-1%7D%5Cdisplaystyle%5Cprod_%7B%5Cell%3D1%7D%5E%7Bk-1%7D%283%5Cell-1%29%7D%7B%283k-2%29%21%7Dx%5E%7B3k-2%7D%7D_%7Bn%3D1%2C4%2C7%2C10%2C%5Cldots%7D)
To show this solution is sufficient, I've attached is a plot of the solution taking
![y(0)=a_0=1](https://tex.z-dn.net/?f=y%280%29%3Da_0%3D1)
and
![y'(0)=a_1=0](https://tex.z-dn.net/?f=y%27%280%29%3Da_1%3D0)
, with
![n=6](https://tex.z-dn.net/?f=n%3D6)
. (I was hoping to be able to attach an animation that shows the series solution (orange) converging rapidly to the exact solution (blue), but no such luck.)
Answer:
Unlike fractions: Fractions with different denominators are called, unlike fractions.
In case of two equivalent fractions, the product of the numerator of one fraction and denominator of the second fraction is equal to the product of the denominator of the first fraction and numerator of the second fraction.
A fraction is said to be in form if its numerator and denominator are 5. To add or subtract like fractions, we add or subtract the keeping the same.
Step-by-step explanation:
4 POINT I DONT NO