# Tag Archives: Evidence

## Evidence of recurrence

To evidence by recurrence, follow the steps below exactly as shown, in the given order:

Step # 1:

Show this is true for n = 1, n = 2…

Step # 2:

Suppose that this is true for n = k

Step # 3:

To prove this is true for n = k + 1

Notes the explanations on evidence by recurrence and important:

Step # 1, you try to show that this is true for specific values. You are free to do this test with a single value or fifty values of your choice or more.

However, showing this is true for the values of one million or more does not prove it will be true for all values. It is a very important observation!

In step # 2, since you have already shown that this is true for one or more values, it is logical to assume or assumes that this is true for n = k or generally.

We usually use the asumption that here we complement or prove step # 3

Step # 3, finally show you that it is true for all values. Note that this step # 2 does not show that this is true for all values.

Example:

Show that for any n, 2 + 4 + 6 +… + 2n = n (n + 1)

Step # 1:

Display the equation is true for n = 1, n = 2…

There is a pitfall to avoid here.

n = 1 means the first value of the expression on the left side. In this case 2

n = 2 means the first two values of the expression on the left side. In this case 2 + 4.

n = 3 means the first three values in the expression on the left side. In this case 2 + 4 + 6

Thus, showing the equation 2 + 4 + 6 +… + 2n = n (n + 1) is true for n = 4 means we need to show that 2 + 4 + 6 + 8 = 4 (4 + 1)

2 + 4 + 6 + 8 = 6 + 6 + 8 = 12 + 8 = 20 and 4 (4 + 1) = 4 × 5 = 20

Since the left side is equal to the right (20 = 20), step # 1 is done. It is not necessary to choose other values that you could do it just for fun and to prove to yourself that it will work for other values.

Step # 2:

Assume that the equation is true for n = k

Just replace n by k.

2 + 4 + 6 +… + 2 k = k (k + 1)

Step # 3:

To prove the equation is true for n = k + 1

It is the most difficult part of the evidence by recurrence. Things can get really hard here. Step in this well problem!

At this stage, you need to write what it means for the equation be true for n = k + 1

Be careful! Just because you wrote what means does not mean that you have proved it. There is another pitfall to avoid when you work on a proof by recurrence.

What this means:

After replacing k by k + 1, you get:

2 + 4 + 6 +… + 2 × (k + 1) = k + 1 (k + 1 + 1)

2 + 4 + 6 +… + 2 × (k + 1) = k + 1 (k + 2)

2 + 4 + 6 + … + 2 × ( k + 1) = ( k + 1 ) × ( k + 2)

We will give you a summary because you may have lost tract of what we are trying to do here.

We do not prove anything yet. The equation 2 + 4 + 6 +… + 2 × (k + 1) = (k + 1) × (k + 2) is what it means for the equation be true for n = k + 1

We are now ready to complete the proof by recurrence using the hypothesis in step no. 2.

on the assumption, 2 + 4 + 6 +… + 2 k = k (k + 1)

Say to yourself, “that the next term is seeking as”?

Since the latter term is now 2 k, the next term should be 2 × (k + 1)

Add 2 × (k + 1) on both sides of the hypothesis

2 + 4 + 6 +… + 2 k + 2 × (k + 1) = k (k + 1) + 2 × (k + 1)

= k2 + k + 2 k + 2.

= k2 + 3 k + 2.

Since 2 = 1 × 2 and 1 + 2 = 3,

k2 + 3k + 2 = ( k + 1) × ( k + 2)

Therefore, 2 + 4 + 6 +… + 2 k + 2 × (k + 1) = (k + 1) × (k + 2) and evidence of recurrence is complete!

The above is a solid and well explained by recurrence. Although study!

## Evidence of base angles theorem

The base angles theorem says that if the sides of a triangle are congruent (isosceles triangle) then the angles opposite those sides are congruent

Start with the following isosceles triangle. The two equal parts are presented with a red mark and contrary to these sides angles are also shown in red

The strategy is to draw the perpendicular bisector of vertex c segment AB

Then use SAS postulate to demonstrate that the trained two triangles are congruent

If the two triangles are congruent, the corresponding angles to be congruent

Draw the perpendicular bisector of c

Since the angle c is cut in two,

angle (x) = angle (y)

Segment AC = segment BC (it has been given)

FC = FC segment segment (on the side of the commune is the same for the triangle ACF and triangle BCF)

Triangles ACF and triangle BCF are consistent and then by SAS or side angle

In other words, by

AC-angle (x) – FC

and

BC-angle (y) – FC

Since the limit ACF and BCF triangle are congruent, angle angle = B

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