Question

已知正項數(shù)列{a_n},其前n項和S_n滿足10S_n=a_n²+5a_n+6，且a₁，a₃，a₁₅成等比數(shù)列，求數(shù)列{a_n}的通項a_n.

Answer 1

解:∵10S_n=a_n²+5a_n+6,                         ①

∴10a₁=a₁²+5a₁+6,解之得a₁=2或a₁=3.

又10S_n-1=a_n-1²+5a_n-1+6(n≥2),           ②

由①-②得10a_n=(a_n²-a_n-1²)+6(a_n-a_n-1),

即(a_n+a_n-1)(a_n-a_n-1-5)=0.

∵a_n+a_n-1＞0,∴a_n-a_n-1=5(n≥2).

當(dāng)a₁=3時,a₃=13,a₁₅=73.

a₁,a₃,a₁₅不成等比數(shù)列,∴a₁≠3.

當(dāng)a₁=2時,a₃=12,a₁₅=72,有a₃²=a₁a₁₅.

∴a₁=2.

∴a_n=5n-3.