当我们遇到有关数列问题有什么简单方法去做题 做好有图示或步骤 或+本人Q229735731

当我们遇到有关数列问题有什么简单方法去做题 做好有图示或步骤 或+本人Q229735731

好麻烦的~~~⊙﹏⊙b汗 打好多字
a(n) = 2×r^(n-1), r 0. n = 1,2,...；r = 1时,S(n) = 2n；r≠1时,S(n) = 2[r^n - 1]/(r-1), n = 1,2,... ；若r = 1,则S(30) = 60 ；S(20) = 40 ；S(10) = 20 . ；2^10S(30) - (2^10 + 1)S(20) + S(10) = 2^10*60 - (2^10 + 1)*40 + 20≠0；所以 r≠1. ；S(30) = 2[r^30 - 1]/(r-1)； S(20) = 2[r^20 - 1]/(r-1)；S(10) = 2[r^10 - 1]/(r-1)；0 = 2^10S(30) - (2^10 + 1)S(20) + S(10)； = 2^10*2[r^30 - 1]/(r-1) - (2^10 + 1)2[r^20 - 1]/(r-1) + 2[r^10 - 1]/(r-1)； 0 = 2^10[r^30 - 1] - (2^10 + 1)[r^20 - 1] + r^10 - 1； = 2^10*r^30 - 2^10 - 2^10*r^20 - r^20 + 2^10 + 1 + r^10 - 1； = 2^10*r^30 - 2^10*r^20 - r^20 + r^10；0 = 2^10*r^20 - 2^10*r^10 - r^10 + 1 ；= 2^10*r^10[r^10 - 1] - [r^10 - 1] ；= [2^10*r^10 - 1][r^10 - 1] ；0 = 2^10*r^10 - 1；r = 1/2. ；a(n) = 2r^(n-1) = (1/2)^n-2, n = 1,2,...； nS(n) = n×2[r^n - 1]/(r-1) = 4n - 4n/2^n.=4（n-n/2^n）设 B(n) = 1/2^1 + 2/2^2 + 3/2^3 + ... + (n-1)/2^(n-1) + n/2^n 2B(n) = 1 + 2/2^1 + 3/2^2 + ... + (n-1)/2^(n-2) + n/2^(n-1), B(n) = 2B(n) - B(n) = 1 + 1/2^1 + 1/2^2 + ... + 1/2^(n-2) + 1/2^(n-1) - n/2^n = [1-1/2^n]/[1-1/2] - n/2^n = 2[1 - 1/2^n] - n/2^n = 2 - (n+2)/2^n, H(n) = 1*S(1) + 2*S(2) + ... + n*S(n) = 1 - 1/2^1 + 2 - 2/2^2 + ... + n - n/2^n = 1 + 2 + ... + n - B(n) = n(n+1)/2 - 2 + (n+2)/2^n
T（n）=4H(n）=2n(n+1) - 8 + 4(n+2)/2^n
哎呀- -你也不加我积分吗