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:
Hypotenuse = 10 Cm
Step-by-step explanation:
Given:
θ = 60°
Base = 5 cm
Hypotenuse = x cm
Find:
Hypotenuse
Computation:
Cos θ = Base / Hypotenuse
Cos 60° = 5 / Hypotenuse
0.5 = 5 / Hypotenuse
Hypotenuse = 10 Cm
minus 90 degree means that it will go anti clock-wise direction. So, this shape will touch on under the y-axis
1 sum
because of the commutative property of addition
a+b=b+a
we can furthuer expand it
a+b+c=a+c+b=b+a+c=b+c+a=c+b+a=c+a+b
same sum
you can test it with numbers
