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名校入学要求

University of Cambridge

Imperial College London

London School of Economics

University of Warwick

University College London

Durham University

Exams

方添优学原创题库，覆盖英国名校核心入学考试。点击任意考试进入刷题。

ESAT

Engineering & Science Admissions Test

剑桥大学工程与自然科学入学考试

数学1 (必考) · 数学2 · 物理 · 化学 · 生物

2300+题目

TMUA

Test of Mathematics for University Admission

大学数学入学测试

数学应用 · 数学推理

1000+题目

TARA

Test of Academic Reasoning for Admissions

牛津与 UCL 部分专业的入学推理测试

Critical Thinking · Problem Solving · Writing Task

1100+题目

STEP

Sixth Term Examination Paper

剑桥大学数学专项考试

STEP 2 · STEP 3

1500+题目

即将上线

AMC

American Mathematics Competitions

AMC 8 · AMC 10 · AMC 12

SMC / BMO

UK Mathematics Challenge & Olympiad

SMC · BMO 1 · BMO 2

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A-Level Mathematics & Further Maths

Pure · Mechanics · Statistics

IGCSE

Cambridge IGCSE Mathematics

Core · Extended

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1Single Choice (1 pts)

Find the value of

\sum_{r=1}^{99} \frac{1}{r(r+1)}

\dfrac{98}{99}

\dfrac{99}{100}

\dfrac{100}{101}

1

\dfrac{99}{101}

▶ Something wrong with this question?

2Single Choice (1 pts)

Given that

\frac{27^{2x-1} \times 81^x}{9^{3x+2}} = 3^k

, what is

k

in terms of

x

x - 7

2x - 7

4x - 7

x + 7

4x + 7

▶ Something wrong with this question?

Incorrect

Answer: B

Why did I get it wrong?

AI Error Analysis

Why you got it wrong

Youtelescopedcorrectly,butstoppedonebracketearly:youevaluated

1-\tfrac{1}{99}

insteadof

1-\tfrac{1}{100}

Root cause

With

r

runningupto

n

,thetermleftatthetailis

\tfrac{1}{n+1}

,not

\tfrac{1}{n}

Concepts to review

Telescoping sumsMisunderstood
Write the final bracket out in full before you cancel anything.
Summation limitsForgotten
Substitute $r=n$ into the second fraction, not the first.

Want to lock this in? Generate a similar-question practice.

Practise

ExplanationHide ⌃

Step 1. Split the term. Partial fractions on

\frac{1}{r(r+1)}

give

\frac{A}{r}+\frac{B}{r+1}

with

A(r+1)+Br=1

. Putting

r=0

gives

A=1

; putting

r=-1

gives

B=-1

\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}

Step 2. Write the sum out in full. The interior terms cancel in pairs:

\left(1-\tfrac{1}{2}\right)+\left(\tfrac{1}{2}-\tfrac{1}{3}\right)+\left(\tfrac{1}{3}-\tfrac{1}{4}\right)+\cdots+\left(\tfrac{1}{99}-\tfrac{1}{100}\right)

Every

-\tfrac{1}{k}

meets a

+\tfrac{1}{k}

in the next bracket, so only the first and the last term survive.

Step 3. Read off the general result.

\sum_{r=1}^{n}\frac{1}{r(r+1)} = 1 - \frac{1}{n+1} = \frac{n}{n+1}

Step 4. Sanity check on a small case. For

n=3

the sum is

\tfrac{1}{2}+\tfrac{1}{6}+\tfrac{1}{12}=\tfrac{6+2+1}{12}=\tfrac{3}{4}

, and the formula gives

\tfrac{3}{3+1}=\tfrac{3}{4}

. It holds.

Step 5. Substitute. Here

n = 99

, so the sum is

1 - \tfrac{1}{100} = \tfrac{99}{100}

, which is B.

Check the upper limit. The last bracket is

\left(\tfrac{1}{99}-\tfrac{1}{100}\right)

, not

\left(\tfrac{1}{98}-\tfrac{1}{99}\right)

. Stop one bracket early and you get

\tfrac{98}{99}

, which is option A.

▶ Something wrong with this question?

‹ PrevSubmitNext

1Single Choice (1 pts)

Find the value of

\sum_{r=1}^{99} \frac{1}{r(r+1)}

\dfrac{98}{99}

\dfrac{99}{100}

\dfrac{100}{101}

1

\dfrac{99}{101}

▶ Something wrong with this question?

Incorrect

Answer: B

AI Error Analysis

Why you got it wrong

Youtelescopedcorrectly,butstoppedonebracketearly:youevaluated

1-\tfrac{1}{99}

insteadof

1-\tfrac{1}{100}

Root cause

With

r

runningupto

n

,thetermleftatthetailis

\tfrac{1}{n+1}

,not

\tfrac{1}{n}

Concepts to review

Telescoping sumsMisunderstood
Write the final bracket out in full before you cancel anything.
Summation limitsForgotten
Substitute $r=n$ into the second fraction, not the first.

Want to lock this in? Generate a similar-question practice.

Practise

ExplanationHide ⌃

Step 1. Split the term. Partial fractions on

\frac{1}{r(r+1)}

give

\frac{A}{r}+\frac{B}{r+1}

with

A(r+1)+Br=1

. Putting

r=0

gives

A=1

; putting

r=-1

gives

B=-1

\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}

Step 2. Write the sum out in full. The interior terms cancel in pairs:

\left(1-\tfrac{1}{2}\right)+\left(\tfrac{1}{2}-\tfrac{1}{3}\right)+\left(\tfrac{1}{3}-\tfrac{1}{4}\right)+\cdots+\left(\tfrac{1}{99}-\tfrac{1}{100}\right)

Every

-\tfrac{1}{k}

meets a

+\tfrac{1}{k}

in the next bracket, so only the first and the last term survive.

Step 3. Read off the general result.

\sum_{r=1}^{n}\frac{1}{r(r+1)} = 1 - \frac{1}{n+1} = \frac{n}{n+1}

Step 4. Sanity check on a small case. For

n=3

the sum is

\tfrac{1}{2}+\tfrac{1}{6}+\tfrac{1}{12}=\tfrac{6+2+1}{12}=\tfrac{3}{4}

, and the formula gives

\tfrac{3}{3+1}=\tfrac{3}{4}

. It holds.

Step 5. Substitute. Here

n = 99

, so the sum is

1 - \tfrac{1}{100} = \tfrac{99}{100}

, which is B.

Check the upper limit. The last bracket is

\left(\tfrac{1}{99}-\tfrac{1}{100}\right)

, not

\left(\tfrac{1}{98}-\tfrac{1}{99}\right)

. Stop one bracket early and you get

\tfrac{98}{99}

, which is option A.

▶ Something wrong with this question?

为什么选我们

01
考场零意外
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03
做完就知道差在哪
不用等老师批改。交卷秒出分，每道题都有详细解析，薄弱点一目了然。
04
哪里不会练哪里
别再从头到尾刷卷子了。按模块、按知识点精准练习，把时间花在刀刃上。
05
看得见的进步
你的每一次练习都被记录。成绩趋势、薄弱环节、进步曲线，全部用数据驱动备考。
06
原创题，配完整解析
剑桥、牛津硕博团队自己出题、自己写解析。解析讲清楚错在哪一步，而不是只丢一个答案给你。

TEAM

100+ 牛剑 / 常春藤背景师资覆盖 G5 所有专业方向原创题目占比 90%+

Blair

IC Biomedical Engineering

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四年交出 140+ 封 IC Offer，亲手打磨 200+ 份 PS，覆盖 BME、MBE、BTV、Design Engineering 四个方向。

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IC Computing AI&ML, CMU ML

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用工业界和学术界的一手经验带学生，把"我想学 CS"转化成有技术深度的申请故事。

Yichen

Cambridge Mathematics

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深耕 ESAT/STEP 数学模块超过 10 年，主导开发 3000+ 道原创数学题目。

Haoyu

Oxford Physics

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负责 ESAT 物理模块内容开发与面试培训，帮助学生建立完整物理思维。

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