Let
![P(n):\ 1+2+\ldots+n = \dfrac{n(n+1)}{2}](https://tex.z-dn.net/?f=P%28n%29%3A%5C%201%2B2%2B%5Cldots%2Bn%20%3D%20%5Cdfrac%7Bn%28n%2B1%29%7D%7B2%7D)
In order to prove this by induction, we first need to prove the base case, i.e. prove that P(1) is true:
![P(1):\ 1 = \dfrac{1\cdot 2}{2}=1](https://tex.z-dn.net/?f=P%281%29%3A%5C%201%20%3D%20%5Cdfrac%7B1%5Ccdot%202%7D%7B2%7D%3D1)
So, the base case is ok. Now, we need to assume
and prove
.
states that
![P(n+1):\ 1+2+\ldots+n+(n+1) = \dfrac{(n+1)(n+2)}{2}=\dfrac{n^2+3n+2}{2}](https://tex.z-dn.net/?f=P%28n%2B1%29%3A%5C%201%2B2%2B%5Cldots%2Bn%2B%28n%2B1%29%20%3D%20%5Cdfrac%7B%28n%2B1%29%28n%2B2%29%7D%7B2%7D%3D%5Cdfrac%7Bn%5E2%2B3n%2B2%7D%7B2%7D)
Since we're assuming
, we can substitute the sum of the first n terms with their expression:
![\underbrace{1+2+\ldots+n}_{P(n)}+n+1 = \dfrac{n(n+1)}{2}+n+1=\dfrac{n(n+1)+2n+2}{2}=\dfrac{n^2+3n+2}{2}](https://tex.z-dn.net/?f=%5Cunderbrace%7B1%2B2%2B%5Cldots%2Bn%7D_%7BP%28n%29%7D%2Bn%2B1%20%3D%20%5Cdfrac%7Bn%28n%2B1%29%7D%7B2%7D%2Bn%2B1%3D%5Cdfrac%7Bn%28n%2B1%29%2B2n%2B2%7D%7B2%7D%3D%5Cdfrac%7Bn%5E2%2B3n%2B2%7D%7B2%7D)
Which terminates the proof, since we showed that
![P(n+1):\ 1+2+\ldots+n+(n+1) =\dfrac{n^2+3n+2}{2}](https://tex.z-dn.net/?f=P%28n%2B1%29%3A%5C%201%2B2%2B%5Cldots%2Bn%2B%28n%2B1%29%20%3D%5Cdfrac%7Bn%5E2%2B3n%2B2%7D%7B2%7D)
as required
Answer:
3 dollars
Step-by-step explanation:
15 x .2
Because you convert the percentage to a decimal
<em>Answer</em><em>:</em>
<em>2</em><em> </em><em>out</em><em>. </em><em>of</em><em> </em><em>7</em>
<em>Explanation</em><em>:</em>
<em>First</em><em> </em><em>of</em><em> </em><em>all</em><em> </em><em>There</em><em> </em><em>are</em><em> </em><em>2</em><em> </em><em>C</em><em> </em><em> </em><em>in</em><em> </em><em>the</em><em> </em><em>word</em><em> </em><em> </em><em>and</em><em> </em><em>so</em><em> </em><em>it's</em><em> </em><em>going </em><em>to </em><em>be</em><em> </em><em>2</em><em> </em><em>out</em><em> </em><em>of </em><em>the</em><em> </em><em>total</em><em> </em><em>number</em><em> </em><em>of</em><em> </em><em>words </em><em>in</em><em> </em><em>the</em><em> </em><em>given </em><em>word</em>
<em>2</em><em> </em><em>/</em><em> </em><em>7</em><em> </em>
<em>So </em><em>that</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>probab</em><em>ility</em><em> </em>
Lines that are not in the same plane are called _____ lines. A. transversal B. parallel C. skew D. perpendicular
Answer: Skew
hope this helps :)