You are not considering the series in reverse, if it said otherwise, sure post your answer but its wrong.
GG, just replace "a" with "1" and "d" with "n"
Isn't he talking about reversing the sum?You are not considering the series in reverse, if it said otherwise, sure post your answer but its wrong.
Try this:
Give one crucial observation and the rest will fall into place
yea, isnt it the same thing as this? O_OYeah, considering the series in descending order
Which is the same thing as writing
=n+(n-1)+...+(n-n+2)+(n-n+1) )
I guess.yea, isnt it the same thing as this? O_O
ahh i didnt get the 2nd line, but i understood the 3rd line. pretty good proof thereI guess.
The way I thought of it was:
Lets take out all the paranthesis and rearrange our sum:
Notice that there are an n number of of n's in the first part of our sum, then the next part of our sum is simply S(n)-n
So:
Rearrange and simplify
=\frac{1}{2}n(n+1) )
I am getting frustrated as well - there is this one year 9 question I am trying at the moment and for some reason I just cant get it.
is it possible if u can post it so we can take a look
I second this.is it possible if u can post it so we can take a look
thisis it possible if u can post it so we can take a look
)
I will scan it up soon - the problem is it has to be solved using year 9 methods, which kind of complicates things. I have a feeling that I am missing something really obvious or I have made a stupid mistake (given my track record, I would say this is almost a certainty).this
(promise I won't be "devious" and tease you about it
I'll bet it's something to do with shapes or geometry?
What about 8?
Read it again slowly (this is for question 8) - it isnt too difficult, but people usually trip up on one small thing.
