第七题 数列的通向及求和,

考察一般项：√[1+1/k^2 +1/(k+1)^2]=√{[k^2(k+1)^2+(k+1)^2+k^2]/[k^2(k+1)^2]=√[k^2(k+1)^2+k^2+2k+1+k^2] /[k(k+1)] /分母已经开根号了,所以平方没有了=√[k^2(k+1)^2+2k^2+2k+1]/[k(k+1)]=√[k^2(k+1)^2+2k(k+1)+1]/[k(k+1)]=√[k(k+1)+1]^2 /[k(k+1)]=[k(k+1)+1]/[k(k+1)]=1+ 1/[k(k+1)]=1+ 1/k -1/(k+1)√(1+1/1^2+1/2^2)+√(1+1/2^2+1/3^2)+。。。+√(1+1/100^2+1/101^2)=(1+1/1-1/2)+(1+1/2-1/3)+。。。+(1+1/100-1/101)=100+(1-1/2+1/2-1/3+。。。+1/100-1/101)=100+(1-1/101)=100- 100/101=10000/101 /这就是这道题的答案。推广：√(1+1/1^2+1/2^2)+√(1+1/2^2+1/3^2)+。。。+√[1+1/n^2+1/(n+1)^2]=(1+1/1-1/2)+(1+1/2-1/3)+。。。+[1+1/n-1/(n+1)]=100+[1-1/2+1/2-1/3+。。。+1/n-1/(n+1)]=100+[1-1/(n+1)]=101 -100/(n+1)