Proving a relation for all natural numbers involves proving it for n = 1 and showing that it holds for n + 1 if it is assumed that it is true for any n.
The relation 2+4+6+...+2n = n^2+n has to be proved.
If n = 1, the right hand side is equal to 2*1 = 2 and the left hand side is equal to 1^1 + 1 = 1 + 1 = 2
Assume that the relation holds for any value of n.
2 + 4 + 6 + ... + 2n + 2(n+1) = n^2 + n + 2(n + 1)
= n^2 + n + 2n + 2
= n^2 + 2n + 1 + n + 1
= (n + 1)^2 + (n + 1)
This shows that the given relation is true for n = 1 and if it is assumed to be true for n it is also true for n + 1.
<span>By mathematical induction the relation is true for any value of n.</span>
Answer:
The answer will be D. 181
Step-by-step explanation:
603 - 422 = 181
<em>Please mark me as the brainliest.</em>
You can use the Pythagorean Theorem. Which is a^2 + b^2 = c^2.
So now to plug in the equation.
In your case a is 8 and c is 34. So now we need to solve for b.
8 squared is equal to 64.
34 squared is equal to 1156.
Now to plug things in.
64 + b^2 = 1156
Now subtract 64 from 1156.
That equals 1092.
The equation is now
b^2 = 1092
Now do the square root of 1092.
But since the square root is an irrational number you can just write b = 1092 ( but put a square root with 1092)
There is a formula which employs the use of determinants and which helps us calculate the area of a triangle if the vertices are given as 
. The formula is as shown below:
Area=
Now, in our case, we have: 
, and
Thus, the area in this case will become:
Area=
Therefore, Area=![\frac{1}{2}\times [[3(-1\times 1-(-5)\times 1]-3[3\times 1-(-2)\times 1]+1[3\times -5-2]]= \frac{1}{2}\times -20=-10](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5B%5B3%28-1%5Ctimes%201-%28-5%29%5Ctimes%201%5D-3%5B3%5Ctimes%201-%28-2%29%5Ctimes%201%5D%2B1%5B3%5Ctimes%20-5-2%5D%5D%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20-20%3D-10)
We know that area cannot be negative, so the area of the given triangle is <u>10 squared units</u>.
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