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enter image description hereenter image description here Hi, I need help with this. I haven't got a clue how to solve this question. I'm familiar with AP, GP, HP, AGP, some special series... I tried, but couldn't figure out how to approach this question. I have attached two pictures, one is the question and the other one is the answer to the question. Thanks

And, btw, there's something else I don't understand. It's pretty obvious that the general term of this series is t(n) = n(1-x)(1-2x)(1-3x).....[1-(n-1)x] But how is it possible? If we put n = 1, I don't think it produces the first term, which is 1. Likewise, putting n=2 doesn't produce the second term, which is 2(1-x). Am I missing something here?enter image description here

4d_
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    You're wrong. For $n=1$ and $n=2$, it does produce the first two terms. For $n=1$, you should have the product $(1-x)\dots(1-0\cdot x)$ in *increasing order of the coefficients of $x$, hence this product is empty, and conventionallly, an empty product is equal to $1$, just as an empty sum is equal to $0$. – Bernard Jul 02 '18 at 11:28
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    The general term can be written as $n.(1-0x).(1-x).(1-2x).\dots.(1-(n-1)x)$. Thus if $n=1$, this gives $1.(1-0x) = 1$ (that's all since $1-(n-1)x=1-0x=1$). – paf Jul 02 '18 at 11:31
  • I didn't quite get it, this 'empty product'. From what I see, putting n = 1 gives us 1(1-x)(1-2x)(1-3x).......(1+2x)(1+x)(1-0x) – 4d_ Jul 02 '18 at 11:45
  • How is this equal to 1? – 4d_ Jul 02 '18 at 11:45
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    The $n^\text{th}$ term is $n$ times a product of $n-1$ factors of the form $(1-kx)$, where the smallest $k$ is $1$ and the largest $k$ is $n-1$. When $n=2$, you get $2$ times a product of one factor, $(1-x)$, i.e. $2(1-x)$, as expected. For $n=1$, you need to be comfortable with the convention that a product of $0$ terms is $1$. Think about how a sum of zero things is zero; when you add up nothing, the result is nothing. Just like $0$ is the neutral element for addition, $1$ is the neutral element for multiplication. – Mike Earnest Jul 02 '18 at 13:48
  • I have just uploaded the solution to this problem. I don't understand it though. Can you please help me understand what's that bar sign over (r-1) ? It's not the recurring sign, is it? – 4d_ Jul 04 '18 at 06:07
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    It's far from your first post here, and you still keep posting pictures. Would you please learn to use MathJax for formatting the math in your math questions? – zipirovich Jul 05 '18 at 03:49
  • By the way, the bar sign over $r-1$ is the vinculum used here for grouping instead of parentheses. It is VERY surprising, though, too see it in a modern book, alongside with the usual parentheses, switching back and forth between the two notations. According to the Wikipedia article linked above, the vinculum for grouping went out of usage by the eighteenth century. – zipirovich Jul 05 '18 at 04:01
  • Will try to learn it. And btw, no one asked me not to post pictures. Is it against the rules? First I posted the question because I needed to know how to approach it. To me, it's more important to know how to approach a question, rather than just knowing the solution. But no one answered, how to solve this question. That's when I uploaded the solution. – 4d_ Jul 05 '18 at 05:24
  • And before I added this question about the bar sign, I tried asking it in the comments first, but again, no me explained it. Since it was the solution to the same question (that this post is all about), I thought uploading the solution would be ok, because I didn't understand what that bar sign was. And thanks for the explaining what that bar symbol is. Basically, it serves the same purpose as parentheses, from what I understand – 4d_ Jul 05 '18 at 05:24

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