结果1 题目题【目】设 _ 题【目】设 _ \$\frac { 1 } { \sqrt { n } }\$ 相关知识点: 试题来源: 解析 【解析】当n=1,2时 _ ;当 _ 时 【解析】当n=1,2时 _ ;当 _ 时 \$f ( n ) \sqrt { n + 1 }\$ 反馈 收藏
Answer and Explanation:1 Here the given series is: {eq}\displaystyle \sum_{n=1}^{\infty }\frac{1}{\sqrt{n}} {/eq} This means: {eq}f(x) = \frac{1}{\sqrt{x}} {/eq}. It is... Learn more about this topic: Infinite Series & Partial Sums: Explanation, Examples & T...
结果1 题目 \$\frac { \sqrt { n } } { n - 1 } \frac { \sqrt { n } } { n } = \frac { 1 } { \sqrt { n } }\$ 相关知识点: 试题来源: 解析 \$\left[ p = \frac { 1 } { 2 } \right.\$ ≤1]。 反馈 收藏 ...
sqrt是求平方根 frac是分数的意思，frac{m}{n} 就是 m/n 没什么意思，就是把那个函数包含起来。有点像编程的样子。比如上面那个$\frac{p}{2}$=2 其实就是p/2=2 我把符号那些去掉，弄在图片里给你看吧。
Chedid, F.B.: A group quorum system of degree \({1+\sqrt{1+n/m}}\) . In: Proceedings of 8th International Conference on Distributed Computing and Networking (ICDCN), Lecture Notes in Computer Science, vol. 4308, pp. 70–81. Springer (2006)Chedid, F.B.: A Group Quorum System of...
解:∵数列{a{}_{n}}的通项公式是a{}_{n}=\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}, ∵前n项和为10, ∴a{}_{1}+a{}_{2}+…+a{}_{n}=10,即(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+…+\sqrt{n+1}-\sqrt{n}= \sqrt{n+1}-1=10, 解得n=...
解:∵{a}_{n}=\frac{1}{\sqrt{n}+\sqrt{n+1}}, ∴{a}_{n}=\sqrt{n+1}-\sqrt{n}, ∴{S}_{n}= \sqrt{2}−1+\sqrt{3}−\sqrt{2}+…+\sqrt{n+1}−\sqrt{n} =\sqrt{n+1 }−1=9 ∴\sqrt{n+1}=10, ∴n=99 故选:B.反馈...
x→0lim3x+1−1x+1−1=23 Explanation: Let: t=6x+1 Then: x→0lim3x+1−1x+1−1=t→1limt2−1t3−1 ... Ex with lim: x→4lim(x−3−12x−7−1) ? https://socratic.org/questions/ex-wit...
{a}_{n}=\sqrt{n+1}-\sqrt{n},利用“累加求和”即可得出。解:∵{a}_{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}= \sqrt{n+1}−\sqrt{n} , ∴数列{{a}_{n}}的前n项和{S}_{n}=( \sqrt{2} -1)+( \sqrt{3} − \sqrt{2} )+…+(\sqrt{ n+1} −\sqrt{n} ...
由于{a}_{n}=\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n},利用“累加求和”即可得出。解:∵{a}_{n}=\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n},∴{S}_{99}=(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+...+(\sqrt{100}-\sqrt{99})=\sqrt{1...