chore(i8n,learn): processed translations

curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-65-convergents-of-e.md

@@ -1,63 +1,45 @@
 ---
 id: 5900f3ad1000cf542c50fec0
-title: 问题65：e的收敛
+title: 'Problem 65: Convergents of e'
 challengeType: 5
-videoUrl: ''
+forumTopicId: 302177
 dashedName: problem-65-convergents-of-e
 ---
 
 # --description--
 
-2的平方根可以写成无限连续分数。
+The square root of 2 can be written as an infinite continued fraction.
 
-√2= 1 + 1
+$\\sqrt{2} = 1 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2 + ...}}}}$
 
-2 + 1
+The infinite continued fraction can be written, $\\sqrt{2} = \[1; (2)]$ indicates that 2 repeats *ad infinitum*. In a similar way, $\\sqrt{23} = \[4; (1, 3, 1, 8)]$. It turns out that the sequence of partial values of continued fractions for square roots provide the best rational approximations. Let us consider the convergents for $\\sqrt{2}$.
 
-2 + 1
+$1 + \\dfrac{1}{2} = \\dfrac{3}{2}\\\\ 1 + \\dfrac{1}{2 + \\dfrac{1}{2}} = \\dfrac{7}{5}\\\\ 1 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2}}} = \\dfrac{17}{12}\\\\ 1 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2}}}} = \\dfrac{41}{29}$
 
-2 + 1
+Hence the sequence of the first ten convergents for $\\sqrt{2}$ are:
 
-2 + ......
+$1, \\dfrac{3}{2}, \\dfrac{7}{5}, \\dfrac{17}{12}, \\dfrac{41}{29}, \\dfrac{99}{70}, \\dfrac{239}{169}, \\dfrac{577}{408}, \\dfrac{1393}{985}, \\dfrac{3363}{2378}, ...$
 
-可以写出无限连续分数，√2= \[1;（2）]，（2）表示2无限重复。以类似的方式，√23= \[4;（1,3,1,8）]。事实证明，平方根的连续分数的部分值序列提供了最佳的有理近似。让我们考虑√2的收敛。
+What is most surprising is that the important mathematical constant, $e = \[2; 1, 2, 1, 1, 4, 1, 1, 6, 1, ... , 1, 2k, 1, ...]$. The first ten terms in the sequence of convergents for `e` are:
 
-1 + 1 = 3/2
+$2, 3, \\dfrac{8}{3}, \\dfrac{11}{4}, \\dfrac{19}{7}, \\dfrac{87}{32}, \\dfrac{106}{39}, \\dfrac{193}{71}, \\dfrac{1264}{465}, \\dfrac{1457}{536}, ...$
 
-2
+The sum of digits in the numerator of the 10<sup>th</sup> convergent is $1 + 4 + 5 + 7 = 17$.
 
-1 + 1 = 7/5
-
-2 + 1
-
-2
-
-1 + 1 = 17/12
-
-2 + 1
-
-2 + 1
-
-2
-
-1 + 1 = 41/29
-
-2 + 1
-
-2 + 1
-
-2 + 1
-
-2
-
-因此，√2的前十个收敛的序列是：1,3 / 2,7 / 5,17 / 12,41 / 29,99 / 70,239 / 169,577 / 408,1333 / 985,3333 / 2378 ，...最令人惊讶的是重要的数学常数，e = \[2; 1,2,1,1,4,1,1,6,1，......，1,2k，1，......]。 e的会聚序列中的前十个项是：2,3,8 / 3,11 / 4,19 / 7,87 / 32,106 / 39,193 / 71,1264 / 465,1457 / 536，.... ..第10个收敛的分子中的数字之和为1 + 4 + 5 + 7 = 17。求e的连续分数的第100个收敛的分子中的位数之和。
+Find the sum of digits in the numerator of the 100<sup>th</sup> convergent of the continued fraction for `e`.
 
 # --hints--
 
-`euler65()`应该返回272。
+`convergentsOfE()` should return a number.
 
 ```js
-assert.strictEqual(euler65(), 272);
+assert(typeof convergentsOfE() === 'number');
+```
+
+`convergentsOfE()` should return 272.
+
+```js
+assert.strictEqual(convergentsOfE(), 272);
 ```
 
 # --seed--