求等比數列首9項之和?

設首項為 a，公比為 r。

T(5) = √2
ar⁴ = √2 ...... [1]

T(1) × T(2) × T(3) × ....... × T(8) × T(9) = T(9)
T(1) × T(2) × T(3) × ....... × T(8) = 1
a × ar × ar² × ...... × ar⁷ = 1
a⁸r²⁸ = 1 ...... [2]

[1]⁴ :
a⁸r³² = 16 ...... [3]

[3]/[2] :
r⁴ = 16
r = ±2

代入 [1] 中：
a(±2)⁴ = √2
a = (√2)/16

當 r = 2 :
首 9 項之和 S(9)
= a(r⁹ - 1)/(r - 1)
= [(√2)/16] × (2⁹ - 1) / (2 - 1)
= 511(√2)/16

當 r = -2 :
首 9 項之和 S(9)
= a[r⁹ - 1)/(r - 1)
= [(√2)/16] × [(-2)⁹ - 1] / [(-2) - 1]
= 513(√2)/48
= 171(√2)/16

設首項為 a，公比為 r。

T(5) = √2
ar⁴ = √2 ...... [1]

T(1) × T(2) × T(3) × ....... × T(8) × T(9) = T(9)
T(1) × T(2) × T(3) × ....... × T(8) = 1
a × ar × ar² × ...... × ar⁷ = 1
a⁸r²⁸ = 1 ...... [2]

[1]⁴ :
a⁸r³² = 16 ...... [3]

[3]/[2] :
r⁴ = 16
r = ±2

代入 [1] 中：
a(±2)⁴ = √2
a = (√2)/16

當 r = 2 :
首 9 項之和 S(9)
= a(r⁹ - 1)/(r - 1)
= [(√2)/16] × (2⁹ - 1) / (2 - 1)
= 511(√2)/16

當 r = -2 :
首 9 項之和 S(9)
= a[r⁹ - 1)/(r - 1)
= [(√2)/16] × [(-2)⁹ - 1] / [(-2) - 1]
= 513(√2)/48
= 171(√2)/16

由題 有一等比數列 a5=2^(1/2), a1*a2*...*a9=a9 => (2^(1/2))^9(=根號2的九次方 因為a1*a9=a2*a8=a5*a5 所以前九項的積 可用a5表示)=a9=a5*(r^4))=(2^(1/2))*(r^4) (r為公比),
所以(2^(1/2))^9=(2^(1/2))*(r^4)=>(2^(1/2))^8=(r^4)=2^4, 因此r=2 or -2,

這裡討論正的情況 負的依此類推 只是代入而已 所以此數列前九項為 (2^(1/2))/16, (2^(1/2))/8, (2^(1/2))/4, (2^(1/2))/2, (2^(1/2)), (2^(1/2))*2, (2^(1/2))*4, (2^(1/2))*8, (2^(1/2))*16
加起來就是=31*(2^(1/2))+15(2^(1/2))/16