1,a1c1+a2c2+.+ancn=n(n+1)(n+2)q则,
2,a1c1+a2c2+.+ancn+a(n+1)c(n+1)=(n+1)(n+2)(n+3)q
所以2,-1,推出a(n+1)c(n+1)=(n+1)(n+2)(n+3)q-n(n+1)(n+2)q
又{an}=n所以{cn}=(2n+4)q
所以cn为等差数列所以Wn就好求了
1,a1c1+a2c2+.+ancn=n(n+1)(n+2)q则,
2,a1c1+a2c2+.+ancn+a(n+1)c(n+1)=(n+1)(n+2)(n+3)q
所以2,-1,推出a(n+1)c(n+1)=(n+1)(n+2)(n+3)q-n(n+1)(n+2)q
又{an}=n所以{cn}=(2n+4)q
所以cn为等差数列所以Wn就好求了