Proof by Induction: Proving ∑(-1)ii2 = (-1)n n(n+1)/2

In summary, the statement for n ∈ [B]N[B] (the natural numbers), ∑ (-1)ii2 (sum from i=1 to n) = (-1)n n(n+1)/2 can be proven using proof by induction, where P(n) is true for all n ∈ [B]N[B]. The basis step is when n = 1, and the inductive step involves showing that ∑ (-1)ii2 (sum from i=1 to k+1) = (-1)k+1(k+1)^2/2 by using the inductive hypothesis. By factoring out (-1)^(k+1), the statement can be proven for all natural numbers.
  • #1
simmonj7
66
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Homework Statement


For n ∈ N (the natural numbers), ∑ (-1)ii2 (sum from i=1 to n) = (-1)n n(n+1)/2.


Homework Equations



For proof by induction, to show that the statement, P(n) is true for all n ∈ N you must show that P(1) is true and P(k+1) is true whenever P(k) is true.

The Attempt at a Solution



Ok so here is where I am at:
We will use induction on n.
Basis step:
For n = 1, we have ∑ (-1)112 (sum from i=1 to 1) = (-1)112 = -1 and (-1)11(1+1)/2 = -1. So, when n =1, ∑ (-1)ii2 (sum from i=1 to n) = (-1)n n(n+1)/2.
Inductive step:
Suppose ∑ (-1)ii2 (sum from i=1 to k) = (-1)kk(k+1)/2. Then ∑ (-1)ii2 (sum from i=1 to k+1) = ∑ (-1)ii2 (sum from i=1 to k) + (-1)k+1(k+1) = (-1)kk(k+1)/2 + (-1)k+1(k+1) by the inductive hypothesis. So, -1kk(k+1)/2 + 2(-1)k+1(k+1)/2 = (-1kk(k+1) + 2(-1k+1)(k+1))/2 = ((-1k + -1k+1)(k+1)(k+1)+1)) /2.

And here is where I am stuck. Cause right now I have the second part i.e. (k+1)((k+1)+1)/2 right, however I can't see how to get the exponents to become -1k+1

Thanks
 
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  • #2
simmonj7 said:

Homework Statement


For n ∈ N (the natural numbers), ∑ (-1)ii2 (sum from i=1 to n) = (-1)n n(n+1)/2.

Homework Equations



For proof by induction, to show that the statement, P(n) is true for all n ∈ N you must show that P(1) is true and P(k+1) is true whenever P(k) is true.

The Attempt at a Solution



Ok so here is where I am at:
We will use induction on n.
Basis step:
For n = 1, we have ∑ (-1)112 (sum from i=1 to 1) = (-1)112 = -1 and (-1)11(1+1)/2 = -1. So, when n =1, ∑ (-1)ii2 (sum from i=1 to n) = (-1)n n(n+1)/2.
Inductive step:
Suppose ∑ (-1)ii2 (sum from i=1 to k) = (-1)kk(k+1)/2. Then ∑ (-1)ii2 (sum from i=1 to k+1) = ∑ (-1)ii2 (sum from i=1 to k) + (-1)k+1(k+1) = (-1)kk(k+1)/2 + (-1)k+1(k+1) by the inductive hypothesis. So, -1kk(k+1)/2 + 2(-1)k+1(k+1)/2 = (-1kk(k+1) + 2(-1k+1)(k+1))/2 = ((-1k + -1k+1)(k+1)(k+1)+1)) /2.

And here is where I am stuck. Cause right now I have the second part i.e. (k+1)((k+1)+1)/2 right, however I can't see how to get the exponents to become -1k+1

Thanks


Your induction step should be looking at (-1)^k*k*(k+1)/2+(-1)^(k+1)*(k+1)^2. You seem to be missing the "^2" on the (k+1). Try it again. Factor out (-1)^(k+1) first.
 
Last edited:
  • #3
Dang. Now that I see that its so easy. Thanks.
 

1. How does proof by induction work?

Proof by induction is a mathematical technique used to prove that a statement is true for all positive integers. It involves two steps: the base case, where the statement is shown to be true for the first integer, and the inductive step, where it is shown that if the statement is true for one integer, it is also true for the next integer.

2. How is proof by induction used to prove the formula ∑(-1)ii2 = (-1)n n(n+1)/2?

In this case, the base case would involve plugging in the first positive integer (n=1) into the formula to show that it is true. The inductive step would then involve assuming the formula is true for some integer k and showing that it is also true for the next integer, k+1. This would involve substituting k+1 for n in the formula and using the fact that the formula is true for k to prove that it is also true for k+1.

3. What is the significance of the alternating signs (-1)ii2 in the formula?

The alternating signs are crucial to the proof by induction, as they allow for the cancellation of terms in the summation. This is what ultimately leads to the simplified formula of (-1)n n(n+1)/2. Without the alternating signs, the formula would not hold true for all positive integers.

4. Can proof by induction be used for any type of mathematical statement?

Proof by induction is a powerful technique that can be used for a wide range of mathematical statements, as long as they involve positive integers. However, it may not be applicable to all types of statements, such as those involving irrational numbers or complex numbers.

5. Are there any alternative methods for proving this formula besides proof by induction?

Yes, there are other methods for proving this formula, such as using properties of geometric series or using mathematical induction with a different base case. However, proof by induction is often the most straightforward and efficient method for proving this type of formula involving positive integers.

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