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>
Me just so I can ask a question
Answer:
-3
Step-by-step explanation:
Here is a different example.
'm' is the gradient (or slope). We measure the difference between the two x values and two y values to see how each of them changed respectively. We know m = rise/run so we can put in our slope (7) and points to find t.
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