Is there a shorter/faster way to show these two expressions are equal?

Solution 1:

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

Solution 2:

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

Solution 3:

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).