Show that n SIGMA d=1 Phi(d) [n/d] = n(n+1)/2 for any positive integer n.
The n should be on top of the SIGMA, and the d=1 is below... it's a sum.
We can use Thm 7.6 (which states that for each positive integer n, SIGMA d|n Phi(d)=n) and Thm 6.11 to help prove this. I understand the basic ideas, and I have done examples to help understand what they're saying. I just need help using these Thms to prove it. My teacher also told us we would use 1+2+3...+n= n(n+1)/2 in the proof.
Please help if you can... thanks!
The n should be on top of the SIGMA, and the d=1 is below... it's a sum.
We can use Thm 7.6 (which states that for each positive integer n, SIGMA d|n Phi(d)=n) and Thm 6.11 to help prove this. I understand the basic ideas, and I have done examples to help understand what they're saying. I just need help using these Thms to prove it. My teacher also told us we would use 1+2+3...+n= n(n+1)/2 in the proof.
Please help if you can... thanks!