I have a proof that I cannot complete. Have done as far as I can go and am now stuck. The RHS is fine, just no idea about the LHS.
Please see the attachment.
Just not sure how to proceed now.
help very much appreciated.
jacs
Hello jacsYou are OK so far. Now plug the value of $\displaystyle c$ back in, and put $\displaystyle x =1$ as in the first proof:
$\displaystyle \frac{2^{n+1}}{n+1}-\frac{1}{n+1}=^nC_0+\tfrac12^nC_1+...+\tfrac{1}{n+ 1}^nC_n =\sum_{r=0}^n\frac{^nC_r}{r+1}$
$\displaystyle \Rightarrow \frac{2^{n+1}-1}{n+1}=\sum_{r=0}^n\frac{^nC_r}{r+1}$
Now note that if we replace n by $\displaystyle (n+1)$ in the first result, $\displaystyle \sum_{r=1}^n{^nC_r}=2^n-1$, we get $\displaystyle \sum_{r=1}^{n+1}{^{n+1}C_r}=2^{n+1}-1$, and the result follows.
Grandad
Thanks Grandad, I followed your explanation and now i see it, it makes perfect sense. I doubt i would have seen that myself though.
Krahl, I wasnt aware of that property, and am making a note of it as it is likely to come in useful. thanks!
Appreciate your help