Mathematical Induction question

[tcesni]

I need help getting started on using Mathematical Induction with this problem...
so what should i do first?

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

 

[tcesni]

never mind...i found the answer

 

[tacman]

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