Can you clarify or add a picture?
Answer:
According to me,
<h2>48 is the answers</h2>
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>
60/100 / 60 are shaded out of the hundred. Simplified would be 3/5. The decimal would be 0.6.
Answer:0.65
Step-by-step explanation:
Firstly, we convert the fractions into decimals:
6/10=0.6
1/5=0.2
1/4=0.25
We then get : 0.6 - 0.2 + 0.25
0.6-0.2=0.4
0.4+0.25=0.65
