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i) By considering $(1+x+x^2+\cdots+x^n)(1-x)$ show that, if $x\neq 1$, $$1+x+x^2+\cdots+x^n=\frac{(1-x^{n+1})}{1-x}$$

ii) By differentiating both sides and setting $x=-1$ show that $$1-2+3-4+\cdots+(-1)^{n-1}n$$

takes the value $-\frac{n}{2}$ if n is even and the value $\frac{(n+1)}{2}$ if n is odd.

For part i) I just simplified the LHS, divided by $(1-x)$ and got the desired result.

For the next part I found the derivative of both sides, and set $x=-1$ giving me:

$$1-2+3-4+\cdots+(-1)^{n-1}n = \frac{(2)(-1(n+1)(-1)^n)-(1-x^{n+1})(-1)}{4} = \frac{-2(n+1)(-1)^n+1+(-1)^{n+2}}{4}$$

However I'm not understanding the part about n being even and odd. If n is even, does this mean that $n = 2n$ and if it is odd, $n = 2n+1/2n-1$? What would be the next step?

Thanks

Michael Albanese
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salman
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  • $n = 2n$ can only be true when $n = 0$. Anyway, when $n$ is even $(-1)^n = 1$, and when is odd, $(-1)^n = -1$. – Tunococ Aug 30 '13 at 11:33
  • The main importance of $n$ in part ii) being even or odd concerns the sign of $(-1)^n$ and $(-1)^{n+2}$. If $n$ is even, then they are both equal to $1$, and if $n$ is odd, they are equal to $-1$. – Arthur Aug 30 '13 at 11:33
  • @Arthur oh, so if we said n to be even, do I replace the n with 2n? – salman Aug 30 '13 at 11:37
  • No. If $n$ is even, it is of the form $2k$ for some integer $k$, not necessarily $k = n$. For example, $6$ is even and is of the form $2\times 3$, not $2\times 6$. – Michael Albanese Aug 30 '13 at 11:39
  • $-2(n+1)(-1)^n + 1 + (-1)^{n+2} = -2(n+1) + 1 + +1 = -2n$ if n is even, and it equals $2(n+1) +1 -1 = 2(n+1)$ is $n$ is odd. – lokodiz Aug 30 '13 at 11:41
  • @SimonC i got that bit now, is there a mathematical way of indicating that n is even or is odd? A particular way of writing it? – salman Aug 30 '13 at 11:42
  • You can say $n \equiv 0 ; \text{mod} ; 2$ if $n$ is even, or $n \equiv 1 ; \text{mod} ; 2$ if $n$ is odd, but just saying whether it is even or odd explicitly is normally fine. – lokodiz Aug 30 '13 at 11:44
  • Also if n is even, what happens to the (n+1)? Does that not change? – salman Aug 30 '13 at 11:45
  • @user90771 Normally "$n$ is even" suffices, or $n\equiv 0 \pmod 2$ as SimonC says. A more concise option is $2|n$. – not all wrong Aug 30 '13 at 11:45
  • @Sharkos sorry to ask, but what does mod 2 mean? – salman Aug 30 '13 at 11:46
  • @user90771 It means "up to adding multiples of 2". The "mod" is short for "modular" as in modular arithmetic: http://en.wikipedia.org/wiki/Modular_arithmetic – not all wrong Aug 30 '13 at 11:47
  • @Sharkos Is this an advanced concept in maths? I have never come across this. – salman Aug 30 '13 at 11:49
  • It's not advanced in that you'll be familiar with "clock arithmetic", but (at least in the UK) you'll normally be first introduced to it as a first year undergraduate – lokodiz Aug 30 '13 at 11:50
  • @SimonC oh thanks, I'm currently doing a-level maths/further so I was worried as to why I hadn't covered it. With respect to the question, if n is even or odd, why doesn't the $(n+1)$ change? – salman Aug 30 '13 at 11:51
  • The parity of $n$ (i.e. whether it's even or odd) matters for the powers of -1 since $(-1)^n$ is $-1$ for odd $n$ and $1$ for even $n$. However, $n+1$ is still always $n+1$ regardless of whether $n$ is even or odd! – lokodiz Aug 30 '13 at 12:01

1 Answers1

1

If $n$ is even, $n = 2k$ for some integer $k$. Then $(-1)^n = (-1)^{2k} = ((-1)^2)^k = 1^k = 1$ and $(-1)^{n + 2} = (-1)^n\times(-1)^2 = 1\times 1 = 1$. Therefore, we have

\begin{align*} \frac{-2(n+1)(-1)^n + 1 +(-1)^{n+2}}{4} &= \frac{-2(n+1)\times 1 + 1 + 1}{4}\\ &= \frac{-2(n+1) + 2}{4}\\ &= \frac{-2n -2 + 2}{4}\\ &= \frac{-2n}{4}\\ &= -\frac{n}{2}. \end{align*}

Can you follow the steps to do the odd case?

Michael Albanese
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  • yeah I managed to do it for the odd case (I like the way it's laid out, this is what I wanted to know). I still am failing to understand why the $(n+1)$ doesn't change when n is odd and even. If n is even then, $n=2k$ for some integer k, then $(n+1)=2k+1$ and if n is odd, then $n=2k-1$ for some integer k, then $(n+1) = 2k. How does (n+1) not change between even and odd values of n? – salman Aug 30 '13 at 14:34
  • If you substitute $n = 2k$ at the beginning of the process I've written, your final answer will be $-\frac{2k}{2}$ which you can write as either $-k$ or $-\frac{n}{2}$. Likewise, in the odd case, if you replace $n$ by $2k - 1$ the expression becomes $\frac{2k}{2}$ which you can write as either $k$ or $\frac{n+1}{2}$. The manipulation of $(n + 1)$ does not change depending on whether $n$ is odd or even, whereas the behaviour of $(-1)^n$ does. That is why we don't need to write $(n + 1)$ differently for $n$ odd and even, but we do need to make such considerations for $(-1)^n$. – Michael Albanese Aug 30 '13 at 15:20
  • got it thanks, also when we say that $n=2k$ for integer values of k do we include 0 or the negative numbers? Is -2 even? My teacher once told me numbers alternate in an even odd pattern, is this true? Is 0 included in the set of positive integers or natural numbers? – salman Aug 30 '13 at 15:46
  • Yes, $0$ is even as are $-2$, $-4$, $\dots$ Integers do indeed alternate between even and odd. The integer $0$ is not positive or negative. Whether $0$ is a natural number or not is a matter of convention; some people include it, some people don't. – Michael Albanese Aug 30 '13 at 16:00