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

[simmonj7]

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, -1^kk(k+1)/2 + 2(-1)^k+1(k+1)/2 = (-1^kk(k+1) + 2(-1^k+1)(k+1))/2 = ((-1^k + -1^k+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 -1^k+1

Thanks

 

[Dick]

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, -1^kk(k+1)/2 + 2(-1)^k+1(k+1)/2 = (-1^kk(k+1) + 2(-1^k+1)(k+1))/2 = ((-1^k + -1^k+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 -1^k+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.

 

[simmonj7]

Dang. Now that I see that its so easy. Thanks.