数列bn=1/(n(n+1)) 求bn的前N项的和,要有证明过程 Tn=n/(n+1)

oscir10222022-10-04 11:39:542条回答

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东方剑虹共回答了16个问题 | 采纳率87.5%: n=1/(n(n+1))=(1/n)-(1/(n+1))
Tn=(1/1)-(1/2)+(1/2)-(1/3)+(1/3)-(1/4)+...+(1/n)-(1/(n+1))
=1-(1/(n+1))
=n/(n+1); 1年前

shuihan428 共回答了15个问题 | 采纳率80%: bn=1/(n(n+1))=(1/n)-(1/(n+1))
Tn=(1/1)-(1/2)+(1/2)-(1/3)+(1/3)-(1/4)+...+(1/n)-(1/(n+1))
=1-(1/(n+1))
=n/(n+1); 1年前

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已知数列Bn=1/(n^2+2n),其前n项和为Tn,求证Tn xml19761年前4: chenxs0219 共回答了15个问题 | 采纳率93.3%

Bn=1/(n^2+2n)=1/[n(n+2)]=1/2*[1/n-1/(n+2)]
Tn=1/2*(1-1/3)+1/2*(1/2-1/4)+1/2*(1/3-1/5)+1/2*(1/4-1/6)+...+1/2*[1/n-1/(n+2)]
=1/2*[1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+.+1/n-1/(n+2)]
=1/2[1+1/2-1/(n+1)-1/(n+2)]

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