设SN=根号(1+1/1^2+1/2^2)+根号(1+1/2^2+1/3^2).+根号(1+1/2010^2+1/201
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/11/12 04:57:20
设SN=根号(1+1/1^2+1/2^2)+根号(1+1/2^2+1/3^2).+根号(1+1/2010^2+1/2011^2)则不大于S的最大整
数为:2010
数为:2010
根号难以处理,为了消除根号,我们先给每一个根号都减去“1”,这样只需要比较“小量”→→→思路很重要!
SN - 2010 = 根号(1+1/1^2+1/2^2) -1 +根号(1+1/2^2+1/3^2) -1 .
+ 根号(1+1/2010^2+1/2011^2)
=[√(1+1/1^2+1/2^2) -1]+ [√(1+1/2^2+1/3^2) -1] + ..+ [√(1+1/k^2+1/(k+1)^2) -1] + .
=sum求和:
=∑ [√(1+1/k^2+1/(k+1)^2) -1] 式中 k = 1→2010
=分子有理化
=∑[1/k^2+1/(k+1)^2] / {√[1 + 1/k^2 + 1/(k+1)^2] + 1}
=通分然后合并
=∑[k^2 + (k+1)^2 ] / 分母在下面
{k×(k+1)×大根号[k^2×(k+1)^2 + (k+1)^2 + k^2 ] + k^2×(k+1)^2}
=∑[2k(k+1) +1 ] / {k×(k+1)×大根号[k^2×(k+1)^2 + 2k(k+1) + 1] + k^2×(k+1)^2}
=大根号恰好是这个数的“完全平方数” → k(k+1) + 1 →太巧啦
=∑[2k(k+1) +1 ] / {2×k^2×(k+1)^2 + k(k+1)}
令k(k+1)=M 为了任意描述
=∑[2M +1 ] / {2×M^2 + M}
=∑1 / M
=∑1 / [k(k+1)]
=∑{1/k - 1/(k+1)} 式中 k = 1→2010
=1 - 1/2011
总之:
SN - 2010 < 1 小于 1 而 大于0 很重要
结果你自己来吧
SN - 2010 = 根号(1+1/1^2+1/2^2) -1 +根号(1+1/2^2+1/3^2) -1 .
+ 根号(1+1/2010^2+1/2011^2)
=[√(1+1/1^2+1/2^2) -1]+ [√(1+1/2^2+1/3^2) -1] + ..+ [√(1+1/k^2+1/(k+1)^2) -1] + .
=sum求和:
=∑ [√(1+1/k^2+1/(k+1)^2) -1] 式中 k = 1→2010
=分子有理化
=∑[1/k^2+1/(k+1)^2] / {√[1 + 1/k^2 + 1/(k+1)^2] + 1}
=通分然后合并
=∑[k^2 + (k+1)^2 ] / 分母在下面
{k×(k+1)×大根号[k^2×(k+1)^2 + (k+1)^2 + k^2 ] + k^2×(k+1)^2}
=∑[2k(k+1) +1 ] / {k×(k+1)×大根号[k^2×(k+1)^2 + 2k(k+1) + 1] + k^2×(k+1)^2}
=大根号恰好是这个数的“完全平方数” → k(k+1) + 1 →太巧啦
=∑[2k(k+1) +1 ] / {2×k^2×(k+1)^2 + k(k+1)}
令k(k+1)=M 为了任意描述
=∑[2M +1 ] / {2×M^2 + M}
=∑1 / M
=∑1 / [k(k+1)]
=∑{1/k - 1/(k+1)} 式中 k = 1→2010
=1 - 1/2011
总之:
SN - 2010 < 1 小于 1 而 大于0 很重要
结果你自己来吧
计算:/1-根号2/+/根号3-根号2/+/根号3-根号4/./根号2009-根号2010/
设SN=根号(1+1/1^2+1/2^2)+根号(1+1/2^2+1/3^2).+根号(1+1/2010^2+1/201
化简:|1-根号2|+|根号2-根号3|+|根号3-根号4|+.+|根号2011-根号2012|
化简:|1-根号2|+|根号2-根号3|+|根号3-根号4|+.+|根号99-根号100|
|1-根号2|+|根号2-根号3|+|根号3-根号4|+...+|根号99-根号100|
根号1+根号2+.+根号9=?
(1+根号2+根号3)(1+根号2-根号3)
|根号6 - 根号2|+|根号2 - 1 | - |3-根号6|
根号3分之根号12×根号2-根号1/2
(根号10-根号5)(根号2+根号1)
计算:根号18×根号1/2-根号12/根号3
(根号1+根号2+根号3+根号4+根号5+根号6+根号7+根号8+根号9+根号10)÷2