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If $$\sum_{k=1}^{5}\left(-1\right)^{\left(k+1\right)}\left(ka^{2}-k^{2}a\right)=P\cdot a^{2}+Q\cdot a$$compute P and Q.

I managed to solve it (albeit slowly) by solving for each term from k=1 to k=5 and adding them up, but that seems rather inefficient. I imagine there has to be a better way to solve this problem, and if so, could someone explain it?

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The coefficient of $a^2$ is $(1-2+3-4+5).$ The coefficient of $a$ is $-(1-4+9-16+25).$

If $5$ was replaced with $100,$ you might look for something faster.

You can prove by induction that:

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

This gives, since $n=5$ is odd:

$$3a-15$$

If you replace $5$ with $n$ then you get:

$$\sum_{k=1}^n (-1)^{k+1}(ka^2-k^2a)\\=(-1)^{n+1}\left(\left\lfloor \frac {n+1}2\right\rfloor a^2-\frac{n(n+1)}2a\right)$$


Trivial: $P$ is always a divisor of $Q.$

This is because when $n$ is even, $P=-\frac n2.$ When $n$ is odd, then $P=\frac{n+1}2.$

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  • $\begingroup$ Am I wrong in thinking that $\sum_{k=1}^{5}\left(-1\right)^{\left(k+1\right)}(k) = P$ and $\sum_{k=1}^{5}\left(-1\right)^{\left(k+1\right)}(k^{2}) = Q$ ? I had already posted an answer $\endgroup$ – BCLC May 13 at 16:30
  • $\begingroup$ @BCLC You are off by a sign for $Q,$ but my first answer before the edit had $P$ and $Q$ reversed. $\endgroup$ – Thomas Andrews May 13 at 16:33
  • $\begingroup$ Thomas Andrews edited to $-Q$. wait so OP indeed DOES know about manipulation of summation and summands but is wondering how to get precisely $P$ and $Q$? Or doesn't know about manipulation? Or what? $\endgroup$ – BCLC May 13 at 16:50
  • $\begingroup$ @BCLC I'd learned some of these properties, manipulation stuff, etc before, I just couldn't remember. To both of you, thanks for the help! $\endgroup$ – nkenschaft May 14 at 17:20
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    $\begingroup$ Nah, see my profile. Quote: My answers recently can be classified as "overkill." I go in depth. @BCLC $\endgroup$ – Thomas Andrews May 14 at 19:45
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$$\sum_{k=1}^{5}\left(-1\right)^{\left(k+1\right)}\left(ka^{2}-k^{2}a\right)=P\cdot a^{2}+Q\cdot a$$

$$=[\sum_{k=1}^{5}\left(-1\right)^{\left(k+1\right)}(k)]a^{2}-[\sum_{k=1}^{5}\left(-1\right)^{\left(k+1\right)}(k^{2})]a=P\cdot a^{2}+Q\cdot a$$

Am I wrong in thinking that $\sum_{k=1}^{5}\left(-1\right)^{\left(k+1\right)}(k) = P$ and $\sum_{k=1}^{5}\left(-1\right)^{\left(k+1\right)}(k^{2}) = -Q$ ?

Maybe it depends on what $a,P,Q$ are. I believe that $3x^2+4x+7=Ax^2+Bx+C$ for all $x \in \mathbb R$ if and only if $3=A,4=B,7=C$.

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  • $\begingroup$ Typically, if $a$ is not specified, you want equality for all $a,$ or more abstractly, as abstract polynomials. $\endgroup$ – Thomas Andrews May 13 at 16:34
  • $\begingroup$ @ThomasAndrews typically, if $a$ is not specified, then we should clarify with the source (instructor, textbook, etc) ? $\endgroup$ – BCLC May 15 at 11:25
  • $\begingroup$ @ThomasAndrews Ah wait but you're (also) talking in general like polynomial in the ring sense, where we have all those polynomial vs polynomial function things? $\endgroup$ – BCLC May 15 at 11:26

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