- #1

tcesni

- 3

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so what should i do first?

a + (a+d)+(a+2d)+...+[a+(n-1)d] = (n/2)[2a+(n-1)d]

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- Mathematica
- Thread starter tcesni
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In summary, mathematical induction is a method of proving mathematical statements or formulas that hold true for all natural numbers. It is different from other proof methods and is specifically used for proving statements about natural numbers. The steps of a mathematical induction proof involve proving the base case and using an assumption to prove the statement for n+1. It can be used to prove statements about sums, products, sequences, and other mathematical structures. However, it has limitations and may not be applicable to more complex structures. The base case and inductive step must also be clearly defined and logically connected for the proof to be valid.

- #1

tcesni

- 3

- 0

so what should i do first?

a + (a+d)+(a+2d)+...+[a+(n-1)d] = (n/2)[2a+(n-1)d]

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- #2

tcesni

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never mind...i found the answer

- #3

tacman

- 1,874

- 0

First, let's review what mathematical induction is. It is a method of proof used to show that a statement is true for all natural numbers. It involves two steps: the base case and the inductive step.

The base case is when we show that the statement is true for the first natural number, usually 1. In this case, the base case would be when n = 1.

The inductive step is when we assume that the statement is true for some arbitrary natural number k, and then use that assumption to prove that it is also true for k+1. This step is repeated until we can prove that the statement is true for all natural numbers.

Now, let's apply this to the given problem. The statement we want to prove is that the sum of the first n terms of an arithmetic sequence with first term a and common difference d is equal to (n/2)[2a+(n-1)d].

First, we will show that the statement is true for the base case, when n = 1.

When n = 1, the sum of the first term in the sequence is just a.

On the other side of the equation, we have (1/2)[2a+(1-1)d] = (1/2)[2a+0] = a.

Therefore, the statement is true for n = 1.

Next, we will assume that the statement is true for some arbitrary natural number k, and use that assumption to prove that it is also true for k+1.

So, assuming that the statement is true for n = k, we have:

a + (a+d)+(a+2d)+...+[a+(k-1)d] = (k/2)[2a+(k-1)d] (this is the inductive hypothesis)

Now, we want to prove that the statement is also true for n = k+1. This means that we need to show that:

a + (a+d)+(a+2d)+...+[a+kd] = ((k+1)/2)[2a+kd]

We can do this by adding (a+kd) to both sides of the equation in our inductive hypothesis:

a + (a+d)+(a+2d)+...+[a+(k-1)d] + (a+kd) = (k/2)[2a+(k-1)d] + (a+kd)

= (k

Mathematical induction is a method of proving mathematical statements or formulas that hold true for all natural numbers (1, 2, 3, ...). It involves using a base case (usually n = 1) and proving that if the statement holds true for n, it also holds true for n+1.

Mathematical induction is different from other proof methods because it is used specifically to prove statements that hold true for all natural numbers. Other proof methods, such as direct proof or proof by contradiction, are used for more general statements.

The steps of a mathematical induction proof are as follows:

1. Prove the base case (usually n = 1).

2. Assume the statement holds true for some arbitrary natural number n.

3. Use this assumption to prove that the statement also holds true for n+1.

4. Conclude that the statement holds true for all natural numbers based on the principle of mathematical induction.

Mathematical induction is typically used to prove statements involving sums, products, or sequences of natural numbers. However, it can also be used to prove statements about other mathematical structures, such as graphs or sets.

While mathematical induction is a powerful proof technique, it does have limitations. It can only be used to prove statements that hold true for all natural numbers, and it may not be applicable to more complex mathematical structures. Additionally, the base case and inductive step must be clearly defined and logically connected for the proof to be valid.

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