高二数学,求解答
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设{an}公差为q1,{bn}公差为q2
Sn = (a1 + an) * n / 2 = [a1 + a1 + (n-1) * q1] * n/2
Tn = [b1 + b1 + (n-1) * q2] * n/2
Sn/Tn = {[a1 + a1 + (n-1) * q1] * n/2} / {[b1 + b1 + (n-1) * q2] * n/2}
=[2a1 + q1 * (n-1)]/ [2b1 + q2 * (n-1)] = (38n+14)/(2n+1)
a1,b1为常数,实际上这是有无穷多组解的,但只需要令
2a1 + q1 * (n-1)=38n+14
2b1 + q2 * (n-1) = 2n+1
即可,令q1 = 38, q2 = 2
求出a1 = 26, b1 = 3/2
a6 / b7 = (a1 + 5q1) / (b1 + 6q2) = (26 + 38 * 5) / (3/2 + 6 * 2) = 16
Sn = (a1 + an) * n / 2 = [a1 + a1 + (n-1) * q1] * n/2
Tn = [b1 + b1 + (n-1) * q2] * n/2
Sn/Tn = {[a1 + a1 + (n-1) * q1] * n/2} / {[b1 + b1 + (n-1) * q2] * n/2}
=[2a1 + q1 * (n-1)]/ [2b1 + q2 * (n-1)] = (38n+14)/(2n+1)
a1,b1为常数,实际上这是有无穷多组解的,但只需要令
2a1 + q1 * (n-1)=38n+14
2b1 + q2 * (n-1) = 2n+1
即可,令q1 = 38, q2 = 2
求出a1 = 26, b1 = 3/2
a6 / b7 = (a1 + 5q1) / (b1 + 6q2) = (26 + 38 * 5) / (3/2 + 6 * 2) = 16
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