2020 AMC 10A Problem 8

Attempt Problem 8 of the 2020 AMC 10A below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 10A solutions, or check the answer key.

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8.

What is the value of 1+2+34+5+6+78++197+198+199200? \begin{aligned} &1+2+3-4+5+6+7-8\\ &\quad+\cdots\\ &\quad+197+198+199-200? \end{aligned}

9,8009,800

9,9009,900

10,00010,000

10,10010,100

10,20010,200

Answer: B
Concepts:summationpairing and grouping
Difficulty rating: 1060
Video solution:
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Written solution:

Group the terms in blocks of four: (1+2+34)(1+2+3-4) +(5+6+78)+(5+6+7-8) ++\cdots +(197+198+199200)+(197+198+199-200). The jjth block is (4j3)+(4j2)(4j-3)+(4j-2) +(4j1)4j=8j6+(4j-1)-4j=8j-6.

There are 5050 blocks, so the sum is j=150(8j6)=850512\sum_{j=1}^{50}(8j-6)=8\cdot\dfrac{50\cdot51}{2} 650=9900-6\cdot50=9900. Thus, B is the correct answer.

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