数学必修5等差数列解答题

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数学必修5等差数列解答题数学必修5等差数列解答题 数学必修5等差数列解答题(1)an=a1+(n-1)da2.a4=65(a1+d)(a1+3d)=65(1)a1+a5=182a1+4d=1

数学必修5等差数列解答题
数学必修5等差数列解答题
 

数学必修5等差数列解答题
(1)
an=a1+(n-1)d
a2.a4= 65
(a1+d)(a1+3d) = 65 (1)
a1+a5=18
2a1+4d =18
a1+2d=9 (2)
sub (2) into (1)
(9-d)(9+d)=65
d^2 = 16
d =4
a1=1
an = 1+(n-1)4 = 4n-3
(2)
(ai)^2 =a1.a21
= 81
ai =9 = 4i-3
i = 3
(3)
Sn = (2n-1)n
= 2n^2 -n
k=-1
√(Sn +kn) =√2.n
=>{ √(Sn +kn) } 是等差数列