I want to know if the two expressions are equivalent:

  1. $\frac{1}{2}(k+2)(2a+(k+1)b)$
  2. $\frac{1}{2}(k+1)(2a+kb)+(a+(k+1)b)$

My attempt:

First, I decided to start with 2 as 1 looks complicated to me (expanding it is time consuming).

Since 1 has $\frac{1}{2}$ as a factor, I try to express 2 in terms of $\frac{1}{2}$ to get:

$\frac{1}{2}\left[(k+1)(2a+kb)+2a+2(k+1)b \right]$

Since there is a $(k+2)$ in 1, I decided to expand everything in the square bracket and hope that $(k+2)$ is a factor of that expression:

$\frac{1}{2} \left[ k^2b+3b+2ak+4a+2b\right]$

Using polynomial long division to divide the expression in the square brackets by $(k+2)$, I get:

$\frac{1}{2} \left[(k+2) (kb+b+2a) \right]$

Which can be simplified to:

$\frac{1}{2}(k+2)(2a+b(k+1))$, as required.

Is there a more efficient method to solve this kind of question and how will you reason about it? I feel that I am missing some obvious shortcuts/properties since I had to resort to polynomial long division.

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    $\begingroup$ That's probably as good as anything I could come up with. I would probably have just expanded both. As unpleasant as that may be, it doesn't seem like it would be too bad in this problem. $\endgroup$ – Jacob Bond Jun 27 '14 at 4:40

Try this: $$\eqalign{\frac{1}{2}(k+2)(2a+(k+1)b) &=\frac{1}{2}[(k+1)+1][(2a+kb)+b]\cr &=\frac{1}{2}[(k+1)(2a+kb)+(k+1)b+(2a+kb)+b]\cr &=\frac{1}{2}[(k+1)(2a+kb)+(2a+2kb+2b)]\cr &=\frac{1}{2}(k+1)(2a+kb)+(a+kb+b)\cr &=\frac{1}{2}(k+1)(2a+kb)+(a+(k+1)b)\cr}$$

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  • $\begingroup$ That's Amazing! Could you please explain your thought process while solving this problem? None of the manipulations look obvious to me except for the last 2. $\endgroup$ – mauna Jun 30 '14 at 13:53
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    $\begingroup$ I looked at the two factors in your first expression and noticed that they are very similar to the first two factors in your second expression. This gave the first line, and the rest was just simplifying, noting that the $(k+1)(2a+kb)$ is part of the required answer, and therefore I made sure I didn't change it any further. $\endgroup$ – David Jun 30 '14 at 13:58

By inspection, the two expressions are quadratics in $k$. Therefore, if they are equal at three different values of $k$, they are equal at all values of $k$. The easiest values to check are $k=0$, $-1$, and $-2$.

For $k=0$ the expressions become

$${1\over2}(0+2)(2a+(0+1)b)=2a+b$$ and $${1\over2}(0+1)(2a+0b)+(a+(0+1)b)=2a+b$$

For $k=-1$, they become

$${1\over2}(-1+2)(2a+0b)=a$$ and $${1\over2}(0)(2a-b)+a+0b=a$$

And for $k=-2$, they become

$${1\over2}(0)(2a-b)=0$$ and $${1\over2}(-1)(2a-2b)+(a-b)=0$$

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  • $\begingroup$ "... if they are equal at there different values of $k$, they are equal for all values of $k$..." Could you please tell me what theorem this is referring to? $\endgroup$ – mauna Jun 30 '14 at 14:30
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    $\begingroup$ @mauna, it's basically polynomial interpolation -- see en.wikipedia.org/wiki/Polynomial_interpolation for details. You can think of it as a generalization of the familiar "two (distinct) points determine a (unique) line" theorem: three points determine a parabola, four points determine a cubic, and so forth. $\endgroup$ – Barry Cipra Jun 30 '14 at 15:25

I would look at the coefficients of $a$ and $b$ independently, since each expression can be written as $f(k)\cdot a+g(k)\cdot b$ for some $f()$ and $g()$. For $a$, we need to check that $\frac12(k+2)\cdot 2$ $= \frac12(k+1)\cdot2+1$, which is pretty immediate (cancel the $\frac12$ and the factor of $2$ in both expressions); for $b$, we need to check that $\frac12(k+2)(k+1)$ $=\frac12(k+1)(k)+k+1$. Multiplying by $2$, this comes down to checking that $(k+2)(k+1) = (k+1)(k)+2(k+1)$, and a few moments' looking will show that this is true (note that you can cancel $k+1$ algebraically from both sides).

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  • $\begingroup$ Would the downvoter care to explain themselves? $\endgroup$ – Steven Stadnicki Jun 27 '14 at 4:55

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