Rewriting Sums

[burt]

I have the following summation formula:

$$\sum\limits_{j=0}^{n-1} {x \choose j} \frac{a^jb^{x-j}}{(a+b)^x}=\frac12\text{, i.e. }\sum\limits_{j=0}^{n-1} {x \choose j} \frac{q^{x-j}}{(q+1)^x}=\frac12$$ I want to know how I can write out the sums knowing $n=2$. The thing that is getting me stuck is the equals $\frac12$ at the end. My instinct is to do this: $$\frac{q^x}{\left(q+1\right)^x}+x\left(\frac{q^{x-1}}{\left(q+1\right)^x}\right)=1$$ Is this right?

 

[Dr.Peterson]

I have the following summation formula:

$$\sum\limits_{j=0}^{n-1} {x \choose j} \frac{a^jb^{x-j}}{(a+b)^x}=\frac12\text{, i.e. }\sum\limits_{j=0}^{n-1} {x \choose j} \frac{q^{x-j}}{(q+1)^x}=\frac12$$ I want to know how I can write out the sums knowing $n=2$. The thing that is getting me stuck is the equals $\frac12$ at the end. My instinct is to do this: $$\frac{q^x}{\left(q+1\right)^x}+x\left(\frac{q^{x-1}}{\left(q+1\right)^x}\right)=1$$ Is this right?

Why did you change the 1/2 to 1? If you're told something equals 1/2, it stays 1/2 even if you write it differently. It doesn't say 1/2 n.

 

[burt]

Why did you change the 1/2 to 1? If you're told something equals 1/2, it stays 1/2 even if you write it differently. It doesn't say 1/2 n.

Oh! I was thinking you the entire equation twice - totally forgetting the meaning of the summation notation.