How did this particular equation come about? I haven't seen it before in the summation rules index on wikipedia:

$$\sum\limits_{i=1}^{k+1} x_i =\left(\sum\limits_{i=1}^{k} x_i\right)+x_{k+1} $$

Edit: I guess what i meant to say was if there was any sort of proof for this equation.


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    $\begingroup$ Addition is associative $\endgroup$ – TomGrubb Jul 6 '15 at 18:53

Since $$\sum\limits_{i=1}^{k} x_i =x_1 +x_2+\cdots + x_k$$ Then $$\sum\limits_{i=1}^{\color{red}{k+1}} x_i =x_1 +x_2+\cdots + x_k+\color{red}{x_{k+1}}$$ Therefore $$\sum\limits_{i=1}^{\color{red}{k+1}} x_i = \left(\sum\limits_{i=1}^{k} x_i\right)+\color{red}{x_{k+1}}$$

  • $\begingroup$ Thank you. I know it was a simple question but I was still stuck because I'm just dumb like that. :) $\endgroup$ – Turing101 Jul 7 '15 at 0:21
  • $\begingroup$ @Turing101, your welcome. We all have our moments. Just keep practicing. $\endgroup$ – k170 Jul 7 '15 at 0:31

This is a rather straightforward statement:

The sum of the first $k$ terms of a sequence, plus the "$k+1$"st term, is equal to the sum of the first $k+1$ terms of this sequence.

Try not to let complicated-looking notation get in the way of your understanding!

  • $\begingroup$ I can see how that works, but i was using it in mathematical induction, and i kept getting the wrong answers, so i began searching for a proof. Is there one? $\endgroup$ – Turing101 Jul 6 '15 at 18:56
  • $\begingroup$ This statement is more of a definition than something that should be proved, in my opinion. Especially in an induction question, I'm assuming you're trying to prove an expression for $$\sum_{i=1}^k x_i$$ by showing that this expression holds when adding the $k+1$th term. $\endgroup$ – Ashkay Jul 6 '15 at 18:57
  • $\begingroup$ yeah, that's how I tried to view it when I was doing the questions in the text book but apparently the (k+1) term changes according to the power of the what your trying to sum. Basically if (x_i) is squared, then (k+1) is also squared and so on. So I was wondering how I would adjust it according to any summation that I would do induction proofs for. $\endgroup$ – Turing101 Jul 7 '15 at 0:20

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