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!