1996 AJHSME Problems/Problem 16

Problem

$1-2-3+4+5-6-7+8+9-10-11+\cdots + 1992+1993-1994-1995+1996=$

$\text{(A)}\ -998 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 998$

Solution

Put the numbers in groups of $4$ :

$(1-2-3+4)+(5-6-7+8)+(9-10-11+ 12) + \cdots + (1993-1994-1995+1996)$

The first group has a sum of $0$ .

The second group increases the two positive numbers on the end by $1$ , and decreases the two negative numbers in the middle by $1$ . Thus, the second group also has a sum of $0$ .

Continuing the pattern, every group has a sum of $0$ , and thus the entire sum is $0$ , giving an answer of $\boxed{C}$ .

Solution 2

Let any term of the series be $t_n$ . Realize that at every $n\equiv0 \pmod4$ , the sum of the series is 0. For $t_{1996}$ we know $1996\equiv0 \pmod4$ so the solution is $\boxed{C}$ .

~Golden_Phi

Solution 3

Compute the sum of the first and last term, the second and second last term, the third and third last term. We get $(1+1996)+(-2-1995)+(-3-1994)+(4+1993)$ and so on. We see that each pair's sum alternates between $1997$ and $-1997$ .We see that every two consecutive pair cancels out. For example $(1+1996)+(-2-1995)$ gives a result of $0$ . There are a total of $1996$ terms, there are $998$ pairs, and there are $499$ pairs which result in a sum of $1997$ each and $499$ pairs which result in a sum of $-1997$ each. These all cancel out to a sum of $0$ , and thus the entire sum is $0$ , giving an answer of $\boxed{C}$ .

~blankbox

1996 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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1996 AJHSME Problems/Problem 16

Contents

Problem

Solution

Solution 2

Solution 3

See Also