A recipe calls for 3 cups of flour to make 2 batches of dinner rolls. One batch makes 15 rolls. Laura is making dinner for 12 pe
2 answers:
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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 :)
On the other hand, we have:
So from (1) and (2) we have:
Now, let's try to calculate sum
but:
When we join both equation there will be:
So the answer is:
Sorry for my bad english, but i hope it won't be a big problem :)
Answer:
Step-by-step explanation:
25 - 3x = -5(1 - x) - 2x
<u></u>
<u>We have to first get rid of the parenthesis:</u>
==> -5
==> 5x
<u>So now you should have:</u>
25 - 3x = -5 + 5x - 2x
<u>Combine like terms:</u>
25 - 3x = -5 + 5x - 2x
25 - 3x = -5 + 3x
<u>Add 3x to both sides:</u>
25 - 3x = -5 + 3x
+ 3x = + 3x
<u>And you should have:</u>
25 = -5 + 6x
<u></u>
<u>Add 5 to both sides:</u>
25 = -5 + 6x
+5 = +5
30 = 6x
<u>Divide 6 to both sides:</u>
=
5 = x