Question

已知數(shù)列{a_n},其中a₂=6且=n,

(1)求a₁,a₃,a₄;

(2)求數(shù)列{a_n}的通項(xiàng)公式.

Answer 1

解析:(1)∵a₂=6,.

解得a₁=1,a₃=15,a₄=28.

(2)由此猜想a_n=n(2n-1).

下面用數(shù)學(xué)歸納法加以證明:

①當(dāng)n=1時(shí),a₁=1×(2×1-1)=1,結(jié)論正確.

②假設(shè)n=k時(shí)結(jié)論正確,即a_k=k(2k-1).

則當(dāng)n=k+1時(shí),有=k.

∴(k-1)a_k+1=(k+1)a_k-(k+1)=(k+1)·k(2k-1)-(k+1)=(k+1)(2k²-k-1)=(k+1)(2k+1)(k-1).

∵k-1≠0,∴a_k+1=(k+1)［2(k+1)-1］,

即當(dāng)n=k+1時(shí)結(jié)論正確.

由①②,可知{a_n}的通項(xiàng)公式是a_n=n(2n-1).