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

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

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- May 1st 2006, 07:29 PMc_323_hEasy problems
Show that the sum of the first $\displaystyle n$ positive odd integers is $\displaystyle n^2$.

Show that the sum of the first $\displaystyle n$ positive even integers is $\displaystyle n^2+n$ - May 1st 2006, 07:34 PMThePerfectHacker
You have,

$\displaystyle 1+3+5+...+(2n-1)$

Or,

$\displaystyle (2\cdot 0-1)+(2\cdot 1-1)+...+(2\cdot n -1)$

Regroup,

$\displaystyle 2(0+1+2+...+n)-(1+1+...+1)$

Using sum formulas,

$\displaystyle 2\cdot \frac{n(n+1)}{2}-n$

Thus,

$\displaystyle n^2+n-n=n^2$ - May 1st 2006, 07:51 PMc_323_hQuote:

Originally Posted by**ThePerfectHacker**

why $\displaystyle (2\cdot 0-1)$? This evaluates to -1, and we're looking for the positive integers. - May 9th 2006, 10:37 AMAfterShock
(2n-1) for definition of odd; start the summation from 1...infinity.

Or, alternatively, 2n+1, and starting it from 0...infinity. - May 9th 2006, 06:55 PMThePerfectHacker
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. - May 28th 2006, 07:48 AMSoroban
Hello, c_323_h!

Quote:

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

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

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

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

Quote:

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

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

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