You could use perturbation method to calculate this sum. Let's start from:
On the other hand, we have:
So from (1) and (2) we have:
Now, let's try to calculate sum
, but this time we use perturbation method.
but:
![S_{n+1}=\sum\limits_{k=0}^{n+1}k\cdot k!=0\cdot0!+\sum\limits_{k=1}^{n+1}k\cdot k!=0+\sum\limits_{k=0}^{n}(k+1)(k+1)!=\\\\\\= \sum\limits_{k=0}^{n}(k+1)(k+1)k!=\sum\limits_{k=0}^{n}(k^2+2k+1)k!=\\\\\\= \sum\limits_{k=0}^{n}\left[(k^2+1)k!+2k\cdot k!\right]=\sum\limits_{k=0}^{n}(k^2+1)k!+\sum\limits_{k=0}^n2k\cdot k!=\\\\\\=\sum\limits_{k=0}^{n}(k^2+1)k!+2\sum\limits_{k=0}^nk\cdot k!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\\\\\\ \boxed{S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n}](https://tex.z-dn.net/?f=S_%7Bn%2B1%7D%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%2B1%7Dk%5Ccdot%20k%21%3D0%5Ccdot0%21%2B%5Csum%5Climits_%7Bk%3D1%7D%5E%7Bn%2B1%7Dk%5Ccdot%20k%21%3D0%2B%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%2B1%29%28k%2B1%29%21%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%2B1%29%28k%2B1%29k%21%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B2k%2B1%29k%21%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%5Cleft%5B%28k%5E2%2B1%29k%21%2B2k%5Ccdot%20k%21%5Cright%5D%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B1%29k%21%2B%5Csum%5Climits_%7Bk%3D0%7D%5En2k%5Ccdot%20k%21%3D%5C%5C%5C%5C%5C%5C%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B1%29k%21%2B2%5Csum%5Climits_%7Bk%3D0%7D%5Enk%5Ccdot%20k%21%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B1%29k%21%2B2S_n%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7BS_%7Bn%2B1%7D%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B1%29k%21%2B2S_n%7D)
When we join both equation there will be:
![\begin{cases}S_{n+1}=S_n+(n+1)(n+1)!\\\\S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\end{cases}\\\\\\ S_n+(n+1)(n+1)!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\\\\\\\\ \sum\limits_{k=0}^{n}(k^2+1)k!=S_n-2S_n+(n+1)(n+1)!=(n+1)(n+1)!-S_n=\\\\\\= (n+1)(n+1)!-\sum\limits_{k=0}^nk\cdot k!\stackrel{(\star)}{=}(n+1)(n+1)!-[(n+1)!-1]=\\\\\\=(n+1)(n+1)!-(n+1)!+1=(n+1)!\cdot[n+1-1]+1=\\\\\\= n(n+1)!+1](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7DS_%7Bn%2B1%7D%3DS_n%2B%28n%2B1%29%28n%2B1%29%21%5C%5C%5C%5CS_%7Bn%2B1%7D%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B1%29k%21%2B2S_n%5Cend%7Bcases%7D%5C%5C%5C%5C%5C%5C%0AS_n%2B%28n%2B1%29%28n%2B1%29%21%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B1%29k%21%2B2S_n%5C%5C%5C%5C%5C%5C%5C%5C%0A%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B1%29k%21%3DS_n-2S_n%2B%28n%2B1%29%28n%2B1%29%21%3D%28n%2B1%29%28n%2B1%29%21-S_n%3D%5C%5C%5C%5C%5C%5C%3D%0A%28n%2B1%29%28n%2B1%29%21-%5Csum%5Climits_%7Bk%3D0%7D%5Enk%5Ccdot%20k%21%5Cstackrel%7B%28%5Cstar%29%7D%7B%3D%7D%28n%2B1%29%28n%2B1%29%21-%5B%28n%2B1%29%21-1%5D%3D%5C%5C%5C%5C%5C%5C%3D%28n%2B1%29%28n%2B1%29%21-%28n%2B1%29%21%2B1%3D%28n%2B1%29%21%5Ccdot%5Bn%2B1-1%5D%2B1%3D%5C%5C%5C%5C%5C%5C%3D%0An%28n%2B1%29%21%2B1)
So the answer is:
Sorry for my bad english, but i hope it won't be a big problem :)
Given:
A(3,0)
B(1,-2)
C(3,-5)
D(7,-1)
1) reflect across x=-4
essentially calculate the difference between the x=-4 line and Px and "add" it in the other direction to x=-4
A(-4-(3-(-4)),0)=A(-11,0)
B(-4-(1-(-4)),-2)=B(-9,-2)
C(-4-(3-(-4),-5))=C(11,-5)
D(-4-(7-(-4)),-1)=D(-15,-1)
2) translate (x,y)->(x-6,y+8)
A(-3,8)
B(-5,6)
C(-3,3)
D(1,7)
3) clockwise 90° rotation around (0,0), flip the x&y coordinates and then decide the signs they should have based on the quadrant they should be in
A(0,-3)
B(-2,-1)
C(-5,-3)
D(-1,-7)
D) Dilation at (0,0) with scale 2/3, essentially multiply all coordinates with the scale, the simple case of dilation, because the center point is at the origin (0,0)
A((2/3)*3,(2/3)*0)=A(2,0)
B((2/3)*1,(2/3)*-2)=B(2/3,-4/3)
C((2/3)*3,(2/3)*-5)=C(2,-10/3)
D((2/3)*7,(2/3)*-1)=D(14/3,-2/3)
Answer:
The number of students who prefer chocolate milk is 780 .
Step-by-step explanation:
Given as :
The total number of students in the school = 1200
The percentage of students who prefer plain white milk = 35 %
Let the number of students who prefer chocolate milk = x
Now, ∵ The percentage of students who prefer plain white milk = 35 %
∴ The percentage of students who prefer chocolate milk = 100 % - 35 % = 65%
So , As The number of students who prefer chocolate milk = x
Or, 65 % of total number of students in school = x
So, x = 
× 1200
or, x = 
∴ x = 780
So, the number of students who prefer chocolate milk = x = 780
And students who prefer plain white milk = 1200 - x = 1200 - 780 = 420
Hence, The number of students who prefer chocolate milk is 780 . Answer
Step-by-step explanation:
Since AC=BE & DE=DC
then measure angle <CAD=<EBD =X
Then< DAB =90-x
And <ABD= 90-X
then measure angle 1= measure angle 2
