设S1=1+1/(1^2)+1/(2^2),S2=1+1/(2^2)+1/(3^2),S3=1+1/(3^2)+1/(4^2).Sn=1+1/[n^2+1/(n+1)^2].设S=√S1+√S2+√S3+.+√Sn,则S=?(用含n的代数式表示,其中n为正整数)
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/04 06:40:18
![设S1=1+1/(1^2)+1/(2^2),S2=1+1/(2^2)+1/(3^2),S3=1+1/(3^2)+1/(4^2).Sn=1+1/[n^2+1/(n+1)^2].设S=√S1+√S2+√S3+.+√Sn,则S=?(用含n的代数式表示,其中n为正整数)](/uploads/image/z/2068716-12-6.jpg?t=%E8%AE%BES1%3D1%2B1%2F%281%5E2%29%2B1%2F%282%5E2%29%2CS2%3D1%2B1%2F%282%5E2%29%2B1%2F%283%5E2%29%2CS3%3D1%2B1%2F%283%5E2%29%2B1%2F%284%5E2%29.Sn%3D1%2B1%2F%5Bn%5E2%2B1%2F%28n%2B1%29%5E2%5D.%E8%AE%BES%3D%E2%88%9AS1%2B%E2%88%9AS2%2B%E2%88%9AS3%2B.%2B%E2%88%9ASn%2C%E5%88%99S%3D%3F%28%E7%94%A8%E5%90%ABn%E7%9A%84%E4%BB%A3%E6%95%B0%E5%BC%8F%E8%A1%A8%E7%A4%BA%2C%E5%85%B6%E4%B8%ADn%E4%B8%BA%E6%AD%A3%E6%95%B4%E6%95%B0%EF%BC%89)
设S1=1+1/(1^2)+1/(2^2),S2=1+1/(2^2)+1/(3^2),S3=1+1/(3^2)+1/(4^2).Sn=1+1/[n^2+1/(n+1)^2].设S=√S1+√S2+√S3+.+√Sn,则S=?(用含n的代数式表示,其中n为正整数)
设S1=1+1/(1^2)+1/(2^2),S2=1+1/(2^2)+1/(3^2),S3=1+1/(3^2)+1/(4^2).Sn=1+1/[n^2+1/(n+1)^2].设S=√S1+√S2+√S3+.+√Sn,则S=?(用含n的代数式表示,其中n为正整数)
设S1=1+1/(1^2)+1/(2^2),S2=1+1/(2^2)+1/(3^2),S3=1+1/(3^2)+1/(4^2).Sn=1+1/[n^2+1/(n+1)^2].设S=√S1+√S2+√S3+.+√Sn,则S=?(用含n的代数式表示,其中n为正整数)
Sn=1+1/n^2+1/(n+1)^2=(n^4+2n^3+3n^2+2n+1)/(n^2*(n+1)^2)=(n*(n+1)+1)^2/(n^2*(n+1)^2)
故√Sn=√(n*(n+1)+1)^2/(n^2*(n+1)^2)=[n(n+1)+1]/[n(n+1)]
所以:
√S1=1+1-1/2
√S2=1+1/2-1/3
√S3=1+1/3-1/4
.
√Sn=1+1/n-1/(n+1)
s= 1+1-1/2 +1+1/2-1/3 1+1/3-1/4 +1+1/(n(n+1)))=n+[(1-1/2)+(1/2-1/3)+...+(1/n-1/(n+1))]=n+1-1/(n+1)
√S1=1+1/(1×2) √S2=1+1/(2×3) ….√Sn=1+1/(n×(n+1))
S=(1+1+…..+1)+1/(1×2)+1/(2×3)+…+1/(n×(n+1))=n+[1-1/(n+1)]
= n+n/(n+1)
1+1/n²+1/(n+1)²通分时分子不用展开应等于[n(n+1)]²+(n+1)²+n² =[n(n+1)]²+2n(n+1)+1 =[n(n+1)+1]²
S1=1+1/(1^2)+1/(2^2),S2=1+1/(2^2)+1/(3^2),S3=1+1/(3^2)+1/(4^2)......Sn=1+1/[n^2+1/
S=自己算
我也不知道
因为Sn=1+1/n^2+1/(n+1)^2=(n^4+2n^3+3n^2+2n+1)/(n^2*(n+1)^2)=(n*(n+1)+1)^2/(n^2*(n+1)^2)
所以√Sn=√(n*(n+1)+1)^2/(n^2*(n+1)^2)=[n(n+1)+1]/[n(n+1)]
所以S=3/2+7/6+13/12+...+[n(n+1)+1]/[n(n+1)]=(1+1/2)+(1+1/6)+...+(1+1/(n(n+1)))=n+[1/2+1/6+...+1/(n(n+1))]=n+[(1-1/2)+(1/2-1/3)+...+(1/n-1/(n+1))]=n+1-1/(n+1)
(n+1)的平方-1/(n+1)