Please Help.

[Peachyyy]

Hi. I have been asked to prove that:

(0) + (.5) + (1) + ... + ((n/2)-1) = (((n/2)-1)(n-1))/2

Can someone please help??

 

[Unco]

Induction beginning with n=2 will work.

It's easiest just to simplify the LHS of the n=k+1 case and the desired RHS right down to show their equality.

 

[Denis]

I think you can use the fact that:
the nth term = n / 2
sum of 1st n terms = n(n+1) / 4

...and go from there.

 

[soroban]

Hello, Peachyyy!

I have been asked to prove that:

$\displaystyle 0\,+\,\frac{1}{2}\,+\,1\,+\,\cdots\,+\,\left(\frac{n}{2}\,-\,1\right)\;=\;\frac{1}{2}\left(\frac{n}{2}\,-\,1\right)(n\,-\,1)$

What a strangely-worded problem . . . it starts with $\displaystyle n\,=\,2$
$\displaystyle \;\;$and the answer could/should have been written: $\displaystyle \,\frac{1}{4}(n\,-\,1)(n\,-\,2)$

Add -$\displaystyle \frac{1}{2}$ to the beginning of the series
$\displaystyle \;\;$and consider: $\displaystyle \:S\;=\;-\frac{1}{2}\,+\,0\,+\,\frac{1}{2}\,+\,1\,+\,\frac{3}{2}\,+\,2\,+\,\cdots\,+\,\left(\frac{n}{2}\,-\,1\right)$

This is an arithmetic series.
$\displaystyle \;\;$It has: $\displaystyle \;$ first term, $\displaystyle a\,=\,-\frac{1}{2},\:$ common difference, $\displaystyle d\,=\,\frac{1}{2},\:$ and $\displaystyle \,n$ terms.


The sum of the first $\displaystyle n$ terms is: $\displaystyle \:S\;=\;\frac{n}{2}[2a\,+\,(n\,-\,1)d]$

$\displaystyle \;\;$Hence, we have: $\displaystyle \:S\;=\;\frac{n}{2}\left[2\left(-\frac{1}{2}\right)\,+\,(n\,-\,1)\frac{1}{2}\right]\;=\;\frac{n}{4}(n\,-\,3)$


Subtract -$\displaystyle \frac{1}{2}$ from this sum: $\displaystyle \,\frac{n}{4}(n\,-\,3)\,-\,\left(-\frac{1}{2}\right) \;= \; \frac{1}{4}(n\,-\,1)(n\,-\,2)\;$ . . . There!