1/n(n+1)(n+2)求和怎么做
1/n(n+1)(n+2)求和怎么做
1/n(n+1)(n+2)
=1/2[2/n(n+1)(n+2)]
=1/2[(n+2)-n]/n(n+1)(n+2)]
=(1/2)[(n+2)/n(n+1)(n+2)-n/n(n+1)(n+2)]
=(1/2)[1/n(n+1)-1/(n+1)(n+2)]
所以和=(1/2)[1/1*2-1/2*3+1/2*3-1/3*4+……+1/n(n+1)-1/(n+1)(n+2)]
=(1/2)[1/1*2-1/(n+1)(n+2)]
=n(n+3)/[4(n+1)(n+2)]
1/n（n-1）（n+2）怎么求和
分母是n(n+1)(n+2)
1/n(n+1)(n+2)=1/2·2/n(n+1)(n+2)
=1/2·[(n+2)-n]/n(n+1)(n+2)
=1/2·[(n+2)/n(n+1)(n+2)-n/n(n+1)(n+2)]
=1/2·[1/n(n+1)-1/(n+1)(n+1)]
例如：1/1×2×3+1/2×3×4+1/3×4×5
=1/2×(1/1×2-1/2×3)+1/2×(1/2×3-1/3×4)+1/2×(1/3×4-1/4×5)
=1/2×(1/1×2-1/2×3+1/2×3-1/3×4+1/3×4-1/4×5）
=1/2×(1/1×2-1/4×5)
=1/2×9/20
=9/40
为何(1/2)*(1/(n+1))*(1/(n+2))=(1/2)*((1/(n+1))-(1/(n+2)))
这个是相等的啊，(1/n+1)-(1/(n+2)=((n+2)-(n+1))/(n+1)(n+2)=1/(n+1)(n+2)=(1/(n+1))*(1/(n+2))
f(n)=1/(n+1)+1/(n+2)+…+1/(n+2^n) 求f(k+1)-f(k)=？
f(k)=1/(k+1)+1/(k+2)+…+1/(k+2^k),
f(k+1)=1/(k+2)+1/(k+3)+…+1/(k+2^k)+1/(k+2^(k+1))
f(k+1)-f(k)=1/(k+2^(k+1))-1/(k+1)
已知f(n)=1/(n+1)+1/(n+2)+.+1/2n，求f(n+1)-f(n).
f(n)=1/(n+1)+1/(n+2)+...+1/2n
f(n+1)=1/(n+2)+1/(n+3)+...+1/2n+1/(2n+1)+1/2(n+1);
f(n+1)-f(n)=1/(2n+1)+1/2(n+1)-1/(n+1)=1/(2n+1)(2n+2);
跟你的答案一样，你的答案应该没错，参考答案有问题？？
1/n-1/(n+2)求和
通分,变成[(n+2)-n]/n(n+2)=2/n(n+2)。ok，结束。
1/n(n+1)(n+2)=1/2[1/n(n+1)-1/(n+1)(n+2)] 为什么是n与(n+1)结合作一个分母 (n+1)(n+2)结合做另一个
就是化简的结果
1/n(n+1)(n+2)
=1/2*2/n(n+1)(n+2)
=1/2*[(n+2)-n]/n(n+1)(n+2)
=1/2*[(n+2)/n(n+1)(n+2)-n/n(n+1)(n+2)]
=1/2*[1/n(n+1)-1/(n+1)(n+2)]
这样就可以和后一项一级前一项互相抵消了
1/n(n+1)+1/(n+1)(n+2)+……+1/(n+9)(n+10)
1/n(n+1)+1/(n+1)(n+2)+……+1/(n+9)(n+10)
=1/n-1/(n+1)+1/(n+1)-1/(n+2)+……+1/(n+9)-1/(n+10)
=1/n-1/(n+10)
=10/(n(n+10))
计算：1／n(n+1) + 1／(n+1)(n+2) + 1/(n+2)(n+3) +……+ 1/(n+2009)(n+2010)
=1/n-1/(n+1)+1/(n+1)-1/(n+2)+......1/(n+2009)-1/(n+2010)
=1/n-1/(n+2010)
=2010/n.(n+2010)
1/(2∧n+1)+1/(2∧n+2)+.+1/(2∧n+2∧n)>=7/12
简单说一下(应该有n>=2这个条件吧)
主要就是 当n=k时 1/k^2 <[1/(k-1)]*[1/k]=[1/（k-1)]-1/k(简单放缩)
也就是1/2^2 < 1-1/2
1/3^2 < 1/2-1/3
1/4^2 < 1/3-1/4
依次写下去 最后1/n^2 < 1/(n-1)-1/n
