设等比数列的前n项和为sn,a4=a1-9,a5,a3,a4成等差数列1求an的通项公式,2证明对任意k(正数),S(k+2),Sk,S(k+1)成等差数列
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设等比数列的前n项和为sn,a4=a1-9,a5,a3,a4成等差数列
1求an的通项公式,2证明对任意k(正数),S(k+2),Sk,S(k+1)成等差数列
题目应该缺少条件,通过第二问,则应该q≠1
(1)设等比数列的公比为q
∵a5.a3.a4成等差数列
∴2a3=a5+a4
∴2a1q²=a1q^4+a1q^3
即 q²+q-2=0
∴ (q+2)(q-1)=0
∴ q=-2(q=1舍)
代入 a4=a1-9
∴ a1*(-8)=a1-9
∴ a1=1
∴ an=(-2)^(n-1)
(2) Sn=[1-(-2)^n]/(1+2)=[1-(-2)^n]/3
∴ Sk=[1-(-2)^k]/3
S(k+1)=[1-(-2)^(k+1)]/3
S(k+2)=[1-(-2)^(k+2)]/3
∴ S(k+2)+S(k+1)
=[1-(-2)^(k+2)]/3+[1-(-2)^(k+1)]/3
=2/3+2*(-2)^(k+1)/3-(-2)^(k+1)/3
=2/3+(-2)^(k+1)/3
=2[1-(-2)^(k+2)]/3
=2Sk
即2Sk=S(k+2)+S(k+1).
∴ 对任意k(正数),S(k+2),Sk,S(k+1)成等差数列
a5,a3,a4成等差数列
2*a3= a5+a4
=>2*(a1*q^2) =a1*q^4+a1*q^3 (1)
a4=a1-9 =>a1*q^3 =a1-9 (2)
由(1)(2)式得出:a1= -9/7 q= -2
an= -9/7 *(-2)^(n-1)
第二题就套公式证明即可
最后得证2sk=S(k+2)+S(k+1)
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