@
Sane: To prove the sum of the first n terms of an arithmetic series, we use
induction. That means, your argument is somewhat circular...
-----------
@
McFlea:
Do you know how induction works?
It goes like this. There are 3 steps.
The first step is to prove that the statement is true for some stating value n (say, n = 1 in your first problem, since the smallest positive integer is 1).
The second step is to assume that the statement holds for n = k (k is some number).
The third, and also the final step is to use your assumption in Step 2, and prove that the statement is true for n = k + 1.
So if the statement is true for n = 1 (the first step), then from the second and the third step, we can see that it's also true for n = 1 + 1 = 2.
And since the statement is true for n = 2, we can conclude that it's true for n = 2 + 1 = 3.
And since the statement is true for n = 3, it must be true for n = 3 + 1 = 4, and so on...
Hence, it's true for all positive integers.
-----------
Ok, I'll give you an example:
Example:
Prove that:
\sum_{i = 1} ^ n (i) = \frac{n (n + 1)}{2}
Step 1:
Test for n = 1, the LHS is 1, and the RHS is also 1, hence the equation holds for n = 1.
Step 2:
Assume that the equation holds for some n = k, ie:
\sum_{i = 1} ^ k (i) = \frac{k (k + 1)}{2}
Step 3:
Use the assumption:
\sum_{i = 1} ^ k (i) = \frac{k (k + 1)}{2}
to prove that:
\sum_{i = 1} ^ {k + 1} (i) = \frac{(k + 1) (k + 2)}{2}.
We start from what we know:
\sum_{i = 1} ^ k (i) = \frac{k (k + 1)}{2}
Adding (k + 1) to both sides, we have:
\Leftrightarrow \left( \sum_{i = 1} ^ k (i) \right) + (k + 1) = \frac{k (k + 1)}{2} + k + 1
\Leftrightarrow \sum_{i = 1} ^ {k + 1} (i) = \frac{k (k + 1) + 2k + 2}{2}
\Leftrightarrow \sum_{i = 1} ^ {k + 1} (i) = \frac{k ^ 2 + 3k + 2}{2}
\Leftrightarrow \sum_{i = 1} ^ {k + 1} (i) = \frac{(k + 1) (k + 2)}{2}
Hence, the equation holds for n = k + 1. (Q.E.D)
Can you go from here?
Let's see if you can finish the 2 problems above.
