Question

设数列为1,22,333,4444,55555……求前1998项的和

Answer 1

数列1,22,333,4444,55555……的通项为n(10^n-1)/9，前n项和Sn=10/9+2*10²/9+3*10³/9+┈┈┈+n*10^n/9-(1/9+2/9+3/9+┈┈┈+n/9)，设An=10/9+2*10²/9+3*10³/9+┈┈┈+n*10^n/9，10An=10²/9+2*10³+┈┈┈+(n-1)*10^n/9+n*10^(n+1)/9，前式减后式得：-9An=10/9+10²/9+10³/9+┈┈┈+n*10^n/9-n*10^(n+1)/9=10(10^n-1)/9²-n*10^(n+1)/9，An=n*10^(n+1)/9²-10(10^n-1)/9³，Sn=n*10^(n+1)/9²-10(10^n-1)/9³-n(n+1)/18，S(1998)=1998*10^(1999)/9²-10[10^(1998)-1]/9³-221889=[17981*10^(1999)+10]/9³-221889