Exercise 32

Doug

You are right. The exponent of $(-1)$ in the last term of the series in the two final expressions should be $m-1$.
So the final two expressions in my answer and in the book should be:

$\displaystyle\sum^{m-1}_{r=0}\binom{2m}{2r+1}(-1)^r=\binom{2m}{1}-\binom{2m}{3}+\binom{2m}{5}- ... +(-1)^{m-1}\binom{2m}{2m-1}$

and

$\displaystyle\binom{2m}{1}-\binom{2m}{3}+\binom{2m}{5}- ... +(-1)^{m-1}\binom{2m}{2m-1}=2^m\sin(m\pi/2)$

Thanks for pointing this out.

I will add this to the book errata for chapter 1 in the forum

The reason you got the right answer for some examples is because $(-1)^{m+1}$ is equal to $(-1)^{m-1}$.

$(-1)^{m+1}=(-1)^m\cdot (-1)^{+1}={(-1)}^m\cdot {(-1)}^{-1}=(-1)^{m-1}$

Vasco

Doug

Yes it's very important to do the exercises I found. Resist the temptation to rush on to the next chapter. When you've read a chapter you tend to think you have understood it all but when you do the exercises only then do you find out whether you really have.

My answers were all scrutinised by Gary and I did the same for his. That's why there are very few errors in them.

Vasco