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Engineering & Science Admissions Test

Cambridge engineering and natural sciences admissions

Maths 1 (Required) · Maths 2 · Physics · Chemistry · Biology

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TMUA

Test of Mathematics for University Admission

Mathematics aptitude test for university entry

Mathematical Application · Mathematical Reasoning

1,000+questions

TARA

Test of Academic Reasoning for Admissions

Reasoning test for selected Oxford and UCL courses

Critical Thinking · Problem Solving · Writing Task

1,100+questions

STEP

Sixth Term Examination Paper

Cambridge advanced mathematics examination

STEP 2 · STEP 3

1,500+questions

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AMC

American Mathematics Competitions

AMC 8 · AMC 10 · AMC 12

SMC / BMO

UK Mathematics Challenge & Olympiad

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Exit1 / 144OK 0/0

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?

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