已知数列an=n(n+1),bn=(n+1)^2,求证1/(a1+b1)+1/(a2+b2)+1/(a3+b3)+……+1/(an+bn)
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已知数列an=n(n+1),bn=(n+1)^2,求证1/(a1+b1)+1/(a2+b2)+1/(a3+b3)+……+1/(an+bn)
已知数列an=n(n+1),bn=(n+1)^2,求证1/(a1+b1)+1/(a2+b2)+1/(a3+b3)+……+1/(an+bn)
已知数列an=n(n+1),bn=(n+1)^2,求证1/(a1+b1)+1/(a2+b2)+1/(a3+b3)+……+1/(an+bn)
其实知道了放缩式子
1/(an+bn)=1/(n+1)(2n+1)< 1/2n(n+1)=1/2*(1/n-1/n+1)
后就大功告成了:
1/(a1+b1)+1/(a2+b2)+1/(a3+b3)+……+1/(an+bn) 把1/(a1+b1)单独拿出来
=1/(2+4)+1/2*(1/2-1/n+1)=1/6+1/4-1/2*1/(n+1)
(an+bn)=n(n+1)+(n+1)^2
=(n+1)(2n+1)>2n(n+1)
1/(an+bn)=1/(n+1)(2n+1)<1/2n(n+1)<(1/2)(1/n-1/(n-1)
从第二项开始放缩,第一项不变,1/(a1+b1)=1/(2+4)=1/6
即1/(an+bn)=1/(n+1)(2n+1)< 1/2n(n+1)=1/2(1/n-...
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(an+bn)=n(n+1)+(n+1)^2
=(n+1)(2n+1)>2n(n+1)
1/(an+bn)=1/(n+1)(2n+1)<1/2n(n+1)<(1/2)(1/n-1/(n-1)
从第二项开始放缩,第一项不变,1/(a1+b1)=1/(2+4)=1/6
即1/(an+bn)=1/(n+1)(2n+1)< 1/2n(n+1)=1/2(1/n-1/n+1)
1/(a1+b1)+1/(a2+b2)+1/(a3+b3)+……+1/(an+bn)<1/6+1/2(1/2-1/3+1/3-1/4+......+1/n-1/(n+1))
=1/6+1/2(1/2-1/(n+1))<1/6+1/4=5/12
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