1. Easy problems

Show that the sum of the first $n$ positive odd integers is $n^2$.

Show that the sum of the first $n$ positive even integers is $n^2+n$

2. You have,
$1+3+5+...+(2n-1)$
Or,
$(2\cdot 0-1)+(2\cdot 1-1)+...+(2\cdot n -1)$
Regroup,
$2(0+1+2+...+n)-(1+1+...+1)$
Using sum formulas,
$2\cdot \frac{n(n+1)}{2}-n$
Thus,
$n^2+n-n=n^2$

3. Originally Posted by ThePerfectHacker
You have,
$1+3+5+...+(2n-1)$
Or,
$(2\cdot 0-1)+(2\cdot 1-1)+...+(2\cdot n -1)$
oops, i meant to post this is the calculus section, but i guess it can apply to number theory.

why $(2\cdot 0-1)$? This evaluates to -1, and we're looking for the positive integers.

4. (2n-1) for definition of odd; start the summation from 1...infinity.

Or, alternatively, 2n+1, and starting it from 0...infinity.

5. You can easily prove these two identities with
Mathematical Induction

I was just too lazy to use induction on these two problems. And decided to use a more elegant way.

6. Hello, c_323_h!

Show that the sum of the first $n$ positive odd integers is $n^2$.
We have: . $1 + 3 + 5 + \hdots + (2n-1)$

. . an Arithmetic Progression with first term $a = 1$ and common difference $d = 2$.

The sum of the first $n$ term is an A.P. is: . $S_n\;=\;\frac{n}{2}[2a + d(n-1)]$

So we have: . $S\;=\;\frac{n}{2}[2\cdot1 + 2(n-1)]\;=\;n^2$

Show that the sum of the first $n$ positive even integers is $n^2+n$
We have another A.P. . $2 + 4 + 6 + \hdots + 2n$

. . with first term $a = 2$ and common difference $d = 2.$

The sum is: . $S\;=\;\frac{n}{2}[2\cdot2 + 2(n-1)] \;= \;n(n + 1)$