根号下1一直加到根号下n怎么求和 相关知识点: 试题来源: 解析 有恒等式:(1^k+2^k+3^k+4^k+...n^k)/n^(k+1)=1/(k+1);所以:(1^0.5+2^0.5+3^0.5+4^0.5+...n^0.5)/n^(0.5+1)=1/(0.5+1);得出:(1^0.5+2^0.5+3^0.5+4^0.5+...n^0.5)=1/(0.5+1)*n^(0.5+1);得出:(1^...
有恒等式：(1^k+2^k+3^k+4^k+...n^k)/n^(k+1)=1/(k+1);所以：(1^0.5+2^0.5+3^0.5+4^0.5+...n^0.5)/n^(0.5+1)=1/(0.5+1);得出：(1^0.5+2^0.5+3^0.5+4^0.5+...n^0.5)=1/(0.5+1)*n^(0.5+1);得出：(1^0.5+2^0.5+3^0.5+4...
根号下1一直加到根号下n怎么求和 答案 有恒等式:(1^k+2^k+3^k+4^k+.n^k)/n^(k+1)=1/(k+1);所以:(1^0.5+2^0.5+3^0.5+4^0.5+.n^0.5)/n^(0.5+1)=1/(0.5+1);得出:(1^0.5+2^0.5+3^0.5+4^0.5+.n^0.5)=1/(0.5+1)*n^(0.5+1);得出:(1^0.5+2^0.5+3^0.5+4^0.5+.n.....
有恒等式:(1^k+2^k+3^k+4^k+.n^k)/n^(k+1)=1/(k+1);所以:(1^0.5+2^0.5+3^0.5+4^0.5+.n^0.5)/n^(0.5+1)=1/(0.5+1);得出:(1^0.5+2^0.5+3^0.5+4^0.5+.n^0.5)=1/(0.5+1)*n^(0.5+1);得出:(1^0.5+2^0.5+3^0.5+4^0.5+.n... 解析看不懂?免费查看同类题视频解析...
有恒等式:(1^k+2^k+3^k+4^k+.n^k)/n^(k+1)=1/(k+1);所以:(1^0.5+2^0.5+3^0.5+4^0.5+.n^0.5)/n^(0.5+1)=1/(0.5+1);得出:(1^0.5+2^0.5+3^0.5+4^0.5+.n^0.5)=1/(0.5+1)*n^(0.5+1);得出:(1^0.5+2^0.5+3^0.5+4^0.5+.n... 解析看不懂?免费查看同类题视频解析...