2020 AMC 10A Problems/Problem 8

Revision as of 21:28, 31 January 2020 by Middletonkids (talk | contribs) (Solution)

Problem

What is the value of

$1+2+3-4+5+6+7-8+\cdots+197+198+199-200?$

$\textbf{(A) } 9,800 \qquad \textbf{(B) } 9,900 \qquad \textbf{(C) } 10,000 \qquad \textbf{(D) } 10,100 \qquad \textbf{(E) } 10,200$

Solution

Split the even numbers and the odd numbers apart. If we group every 2 even numbers together and add them, we get a total of $50\cdot (-2)=-100$. Summing the odd numbers is equivalent to summing the first 100 odd numbers, which is equal to $100^2=10000$. Adding these two, we obtain the answer of $\boxed{\text{(B) }9900}$.


We can break this entire sum down into $4$ integer bits, in which the sum is $2x$, where $x$ is the first integer in this bit. We can find that the first sum of every sequence is $4x-3$, which we plug in for the $50$ bits in the entire sequence is $1+2+3+\cdots+50=1275$, so then we can plug it into the first term of every sequence equation we got above $4(1275)-3(50)=4950$, and so the sum of every bit is $2x$, and we only found the value of $x$, the sum of the sequence is $4950\cdot2=\boxed{(B)9900}$. -middletonkids

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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