Answer:
Step-by-step explanation:
Hello, we want to prove that a proposition depending on n, that we can note P(n), is true for any n positive integer greater than 1. We need to follow several steps.
Step 1 - prove P(1)
For n = 1, n(2n+1)=1*3 =3 so we have
3 = 3, which is obviously true.
First step done!
Step 2 - for
we assume P(k) and we need to prove P(k+1)
We assume that 3+7+11+...+(4k-1)=k(2k+1)
so we can write that
3+7+11+...+(4k-1)+(4(k+1)-1)=k(2k+1)+(4k+4-1)=k(2k+1)+4k+3
![=2k^2+k+4k+3\\\\=2k^2+5k+3](https://tex.z-dn.net/?f=%3D2k%5E2%2Bk%2B4k%2B3%5C%5C%5C%5C%3D2k%5E2%2B5k%2B3)
and
(k+1)(2(k+1)+1)=(k+1)(2k+3)
![=k(2k+3)+2k+3\\\\=2k^2+3k+2k+3\\\\=2k^3+5k+3](https://tex.z-dn.net/?f=%3Dk%282k%2B3%29%2B2k%2B3%5C%5C%5C%5C%3D2k%5E2%2B3k%2B2k%2B3%5C%5C%5C%5C%3D2k%5E3%2B5k%2B3)
These two expressions are the same so it means that P(k+1) is true, meaning that
3+7+11+...+(4k-1)+(4(k+1)-1)=(k+1)(2(k+1)+1)
Step 3 - The conclusion
Finally, we have just proved that
3+7+11+...+(4n-1)=n(2n+1) for any n positive integer > 0
Thank you
Answer:
C
Step-by-step explanation:
Do it then, no one is stopping you right now.
Answer:
hol up
Step-by-step explanation:
We can use proportion and cross multiplying to solve this problem
22% ----------24.2
100%----------x
Now we can cross multiply
![x= \frac{100*24.2}{22} =110](https://tex.z-dn.net/?f=x%3D%20%5Cfrac%7B100%2A24.2%7D%7B22%7D%20%3D110)
- its the our number