已知Sn是数列an的前n项和,且an=7s(n-1)+2(n>=2，n属于N+),a1=2

显示全部楼层 · 2011-11-1 13:41:55

∵an = 7S(n - 1) + 2a(n+1) = 7Sn + 2∴a(n+1) - an = 7ana(n+1) = 8an又∵a1 = 2∴an = 2 * 8^(n-1)log2an = log2[2 * 8^(n-1)] = log2[2^(3n-2)] = 3n - 2bn = 1 / (3n - 2)*(3n + 1) = 1/3 * [1/(3n - 2) - 1/(3n + 1)]Tn = 1/3 * [1/1 - 1/4 + 1/4 - 1/7 + ... + 1/(3n - 2) - 1/(3n + 1)]= 1/3 * [1 - 1/(3n + 1)]=n / (3n + 1)∵lim(n-...

千问 · 2011-11-1 13:41:55

an=7s(n-1)+2 ---1a(n+1)=7sn+2---2联立1、2式并结合an=sn-s(n-1)可得：
a(n+1)-an=7an即：a(n+1)/an=8所以an是首相为2，公比为8的等比数列。an=2*8^(n-1)因为an=2*8^(n-1)=2*2^(3n-3)=2^(3n-2)log2(...