急,n均为下缀 已知{an}递增的等差数列,a1=2,a2^2=a4+8（1）求数列{an}的通项公式（2）若bn=an+

你稍等,我帮你详细写出来谢谢了设公差为d所以 由题意 a2=2+d, a4=2+3d所以 (2+d)²=2+3d+8所以 d²+d-6=0所以 d=2 或者 d=-3(舍去，因为{an}是递增数列)所以 an=2n由题意 bn = 2n+2^(2n) = 2n+4^n所以 Sn = 2+4^1+4+4^2+6+4^3+...+2n+4^n = 4^1+4^2+4^3+...+4^n+(2+4+6+...+2n)=4*(4^n -1)/(4-1) +(2+2n)n/2=(4/3)*(4^n -1) +n(n+1)检验：an=2,4,6,8,10......从而 2^an=4,16,64,256,1024......从而 bn=6,20,70,264,1034......从而 Sn=6,26,96,360,1394......若代入Sn表达式计算：S1 = (4/3)*(4-1) +1*2 = 4+2 = 6S2 = (4/3)*(16-1) +2*3 = 20+6 = 26S3 = (4/3)*(64-1) +3*4 = 84+12 = 96S4 = (4/3)*(256-1) +4*5 = 340+20 = 360S5 = (4/3)*(1024-1) +5*6 = 1364+30 = 1394...... ...... ......可知答案是正确的非常感谢