Question

这个题怎么做，给个详细解析

Answer 1

(1)an^2-2nan-(2n+1)=0,
(an+1)(an-2n-1)=0,an>0,
所返源游以an=2n+1.
(2)bn=(2n+1)*2^n,它的前n项裂弊和
Sn=3*2+5*2^2+……漏销+(2n+1)*2^n,(1)
2Sn=.......3*2^2+……+（2n-1）*2^n+(2n+1)*2^(n+1),(2)
(2)-(1),得Sn=-3*2-2(2^2+2^3+……+2^n)+(2n+1)^(n+1)
=-6-2[2^(n+1)-4)+(2n+1)*2^(n+1)
=(2n-1)*2^(n+1)+2.

Answer 2

sin^4x+cos^4x=（sin^2x+cos^2x）^2-2sin^2xcos^2 =1-2sin^2xcos^2 代入后用sin2x=2tanx/（1+tan^2x）代换即可中间的看掘郑不念基清判高颂就不算了