I am trying to visualize the concept that should be involved with this problem: What is $x$ in the equation $$(x-1)-2(x-2)+3(x-3)-4(x-4)+\dots-10(x-10)=0\quad ?$$
What should I do? Express in summation (which I find very hard) or just algebra?
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$$\sum\limits_{i=1}^{10}i(x-i)=\sum\limits_{i=1}^{10}ix-i^2=x\sum\limits_{i=1}^{10}i-\sum\limits_{i=1}^{10}i^2=\frac{x\cdot 10\cdot11}2-\frac{10\cdot11\cdot21}6=0\to55x=385$$ – Rushabh Mehta Sep 21 '19 at 16:33
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Your sum is $$\sum_{r=1}^{10}r(x-r)$$ – Rishi Sep 21 '19 at 16:33
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the sign changes every term which would denote that (-1) is raised to the power of n (referring to the number of terms) then again multiplied to(-1). That would raise another question. – Pisting Inatay Sep 22 '19 at 12:38
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Note that by using Gauss' trick $$\begin{align}(x-1)-10(x-10)&=-9(x-11)\\ -2(x-2)+9(x-9)&=7(x-11)\\ 3(x-3)-8(x-8)&=-5(x-11)\\ -4(x-4)+7(x-7)&=3(x-11)\\ 5(x-5)-6(x-6)&=-(x-11) \end{align}$$ and each term is zero for $x=11$.
Robert Z
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I like this a lot. Also noting that adding the next term [which would be +11(x--11)] maintains the factor $x-11$. – Mark Bennet Sep 21 '19 at 16:56
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$$(x-1)-2(x-2)+3(x-3)-4(x-4)+\dots-10(x-10)=0$$ $$\implies x(1-2+3-4+\cdots-10)-(1-2^2+3^2-\cdots-10^2)=0$$ $$\implies x=\dfrac{1-2^2+3^2-\cdots-10^2}{1-2+3-4+\cdots-10}$$ $$\implies x=11$$
$$1-2^2+3^2-\cdots-10^2=\sum_{n=1}^{10}(-1)^{n-1}n^2=-55$$ and $$1-2+3-4+\cdots-10=\sum_{n=1}^{10}(-1)^{n-1}n=-5$$
nmasanta
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One thing you can do with problems like this is look for helpful patterns. There is clearly a pattern involved, but does it simplify into something you can work with. The first few terms come out as
$$x-1=0$$$$-x+3=0$$$$2x-6=0$$$$-2x+10=0$$$$3x-15=0$$$$-3x+21=0$$Does that help you to see what is going on.
Mark Bennet
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