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^0.5)=2/3*n^3/2;
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^0.5)=2/3*n^3/2;