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Telescoping: all problems

Focus mode: every practice problem from the chapter, one after another.

0 / 12 solved

Problem 16.6.1

AMC 8

What is the value of 4⋅(−1+2−3+4−5+6−7+⋯+1000)4 \cdot (-1 + 2 - 3 + 4 - 5 + 6 - 7 + \cdots + 1000)?

Problem 16.6.2

AMC 8

What is the value of the expression 1⋅2⋅3⋅4⋅5⋅6⋅7⋅81+2+3+4+5+6+7+8\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}?

Problem 16.6.3

AMC 8

What is the value of the product (1+11)⋅(1+12)⋅(1+13)⋅(1+14)⋅(1+15)⋅(1+16)\left(1+\frac{1}{1}\right) \cdot \left(1+\frac{1}{2}\right) \cdot \left(1+\frac{1}{3}\right) \cdot \left(1+\frac{1}{4}\right) \cdot \left(1+\frac{1}{5}\right) \cdot \left(1+\frac{1}{6}\right)?

Problem 16.6.4

AMC 8

Find the value of the expression 100−98+96−94+92−90+⋯+8−6+4−2100 - 98 + 96 - 94 + 92 - 90 + \cdots + 8 - 6 + 4 - 2.

Problem 16.6.5

AMC 8

What is the value of the product (1⋅32⋅2)(2⋅43⋅3)(3⋅54⋅4)⋯\left(\frac{1 \cdot 3}{2 \cdot 2}\right)\left(\frac{2 \cdot 4}{3 \cdot 3}\right)\left(\frac{3 \cdot 5}{4 \cdot 4}\right) \cdots (97⋅9998⋅98)(98⋅10099⋅99)\left(\frac{97 \cdot 99}{98 \cdot 98}\right)\left(\frac{98 \cdot 100}{99 \cdot 99}\right)?

Problem 16.6.6(additional)

AMC 8

What is the value of 1+3+5+⋯+2017+2019−2−4−6−⋯−2016−20181 + 3 + 5 + \cdots + 2017 + 2019 - 2 - 4 - 6 - \cdots - 2016 - 2018?

Problem 16.6.7(additional)

AMC 8

What is the product of 32×43×54×⋯×20062005\frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \cdots \times \frac{2006}{2005}?

Problem 16.6.8(additional)

AMC 8

What is the value of 13⋅24⋅35⋯1820⋅1921⋅2022\frac{1}{3} \cdot \frac{2}{4} \cdot \frac{3}{5} \cdots \frac{18}{20} \cdot \frac{19}{21} \cdot \frac{20}{22}?

Problem 16.6.9(additional)

MATHCOUNTS

If n=102020−102019+102018−102017+⋯+102−101n = 10^{2020} - 10^{2019} + 10^{2018} - 10^{2017} + \cdots + 10^2 - 10^1 what is the sum of digits of the integer nn?

Problem 16.6.10(additional)

Evaluate 12×3+13×4+⋯+110×11\frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \cdots + \frac{1}{10 \times 11}

Problem 16.6.11(additional)

Omega Learn

Evaluate 1−2−3+4+5+6−7−8−9−10⋯+90+911 - 2 - 3 + 4 + 5 + 6 - 7 - 8 - 9 - 10 \cdots + 90 + 91 (1 positive consecutive value, 2 negative consecutive values, 3 positive consecutive values, …)

Problem 16.6.12(additional)

Omega Learn

Let the sequence ana_n be defined as an=31+2+3+⋯+na_n = 3^{1+2+3+\cdots+n} for all positive integers nn. Sohil is trying to evaluate the following sequence: 12×23×34×…\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \ldots. Due to his bad handwriting, in the expression above, he accidentally writes each multiplication sign before all fractions of the form anan+1\frac{a_n}{a_n+1} (for example, 3131+1\frac{3^1}{3^1+1}, 3(1+2)3(1+2)+1\frac{3^{(1+2)}}{3^{(1+2)}+1}, etc.) with a plus sign. What is the value of the new expression?