求数列an=1/n2前n项和
\u6570\u5217\uff5ban\uff5d\u4e2dan=1/(9n2-1),\u6c42\u524dn\u9879\u548c\u7684\u516c\u5f0fSn=(-1)^(n+1)*n(n+1)/2
\u8bc1\u660e\uff1a1)\u5f53n=1\u65f6Sn=1\u547d\u9898\u6210\u7acb\uff1b
2)\u5047\u8bbe\u5f53n=k\u65f6\u547d\u9898\u6210\u7acb,\u5373
Sk=(-1)^(k+1)*k(k+1)/2
\u5219\u5f53n=k+1\u65f6,
S(k+1)=Sk+a(k+1)=(-1)^(k+1)*k(k+1)/2+(-1)^(k+2)*(k+1)^2
=-(-1)^(k+2)*k(k+1)/2+(-1)^(k+2)*(k+1)^2
=(-1)^(k+2)[(k+1)^2-k(k+1)/2]
=(-1)^(k+1)*(k+1)(k+2)/2\u547d\u9898\u4e5f\u6210\u7acb
\u7efc\u4e0a1)2),Sn=(-1)^(n+1)*n(n+1)/2
\u5728\u6570\u5217{an} \u4e2d,a1=1,a(n+1)=2an+2^n,\u6c42\u6570\u5217\u7684\u524dn\u9879\u548c
a(n+1)=2an+2^n
\u540c\u9664\u4ee52^n
a(n+1)/2^n=2an/2^n+1
a(n+1)/2^n-an/2^(n-1)=1
\u6240\u4ee5\u6570\u5217\uff5ban/2^(n-1)\uff5d\u4e3a\u4ee51\u4e3a\u516c\u5dee\u7684\u7b49\u5dee\u6570\u5217
a1/2^0=1
an/2^(n-1)=1+(n-1)*1=n
\u6240\u4ee5an = n2^(n-1)
Sn=1*2^0+2*2^1+3*2^2+....+ n2^(n-1)
2Sn= 1*2^1+2*2^2+....+(n-1)2^(n-1)+n2^n
\u75282\u5f0f\uff0d1\u5f0f
Sn=-1-2^1-2^2-....2^(n-1)+n2^n
=-1-[2+2^2+2^3+...+2^(n-1)]+n2^n
=(n-1)2^n+1
但可以逼近
此数列的前n项和＜2 可以用放缩法证明
如有疑问，可追问！
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绛旓細a(n+1)=an²a1=1 a2=1²=1 a3=a2² =1 ...an=1
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绛旓細4^(n-1) * b3 = 5*4^(n-3)4^(n-1) * b2 = 3*4^(n-2)4^(n-1) * b1 = 1*4^(n-1)涓婇潰鐨勭瓑寮忥紝宸﹀彸鍒嗗埆鐩稿姞锛屽乏杈瑰緱 4^(n-1) * Sn 鍙宠竟鏄瓑宸鏁板垪锛堥椤逛负2n-1锛屽叕宸负-2锛変笌绛夋瘮鏁板垪锛堥椤逛负1锛屽叕姣斾负4锛夌浉涔樹箣鍚庣殑鍜屻傞噸鐐规槸姹傝繖涓拰锛屾垜浠厛鎶婂畠璁句负x 鍗 4...
绛旓細2)bn锛(2n-1)/an=(2n-1)*(1/4)^(n-1)Sn=1+3*1/4+5*(1/4)^2+7*(1/4)^3+...+(2n-3)*(1/4)^(n-2)+(2n-1)*(1/4)^(n-1)1/4Sn=1/4+3*(1/4)^2+5*(1/4)^3+...+(2n-5)*(1/4)^(n-2)+(2n-1)*(1/4)^n 3/4Sn=1+2[1/4+(1/4)^2+...
