第七题 数列的通向及求和,
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第七题 数列的通向及求和,
![](http://img.wesiedu.com/upload/8/74/87454e195c166e66ad68898586ce46e8.jpg)
![](http://img.wesiedu.com/upload/8/74/87454e195c166e66ad68898586ce46e8.jpg)
![第七题 数列的通向及求和,](/uploads/image/z/6666243-51-3.jpg?t=%E7%AC%AC%E4%B8%83%E9%A2%98+%E6%95%B0%E5%88%97%E7%9A%84%E9%80%9A%E5%90%91%E5%8F%8A%E6%B1%82%E5%92%8C%2C)
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)
√[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)