Question

設(shè)a為常數(shù),且a₁=1-2a,a_n=3^n-1-2a_n-1(n∈N^*,n≥2),

求證:對(duì)任意n≥2,a_n=［3ⁿ+(-1)^n-1·2ⁿ］+(-1)ⁿ·2ⁿa.

Answer 1

證明:(1)當(dāng)n=2時(shí),由已知a₂=1+4a,等式成立;?

(2)假設(shè)當(dāng)n=k(k≥2)時(shí)等式成立,即a_k=［3^k+(-1)^k-1·2^k］+(-1)^k·2^ka,?

那么a_k+1=3^k-2a_k=3^k-［3^k+(-1)^k-12^k］-(-1)^k2^k+1a??

=［3^k+1+(-1)^k·2^k+1］+(-1)^k+1·2^k+1a,?

也就是說(shuō),當(dāng)n=k+1時(shí),等式也成立.?

根據(jù)(1),(2)可知等式對(duì)任何n∈N^*都成立.