Question

高中数学，求解，谢谢

Answer 1

1、（1）d = (a3-a1)/(3-1) = (-3-1)/(3-1) = -2 ，
所以 an = a1+(n-1)d = 1+(n-1)*(-2) = 3-2n 。
（2）Sk = k(a1+ak)/2 = k(1+3-2k)/2 = k(2-k) = -35 ，
所以 k^2-2k-35 = 0 ，(k+5)(k-7) = 0 ，
所以 k = 7 （舍去 -5）。
2、通项 an = a1+(n-1)d = 3-2n ，
S(k+2)-Sk = a(k+1)+a(k+2) = [3-2(k+1)]+[3-2(k+2)] = -4k = -24 （疑似输入错误），
所以 k = 6 。
3、（1）d = (a6-a3)/(6-3) = (7-4)/(6-3) = 1 ，因此 an = a3+(n-3)d = 4+(n-3) = n+1 。
（2）bn = 1/[a(3n+1)*a(3n+4)] = 1/[(3n+2)(3n+5)] = 1/3*[1/(3n+2)-1/(3n+5)]，
所以 Tn = 1/3*[1/5-1/8+1/8+1/11+......+1/(3n+2)-1/(3n+5)]
= 1/3*[1/5-1/(3n+5)]
= n/[3(3n+5)] 。