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1984 AHSME Problems/Problem 12

Problem

If the sequence $\{a_n\}$ is defined by

$a_1=2$

$a_{n+1}=a_n+2n$

where $n\geq1$.

Then $a_{100}$ equals

$\mathrm{(A) \ }9900 \qquad \mathrm{(B) \ }9902 \qquad \mathrm{(C) \ } 9904 \qquad \mathrm{(D) \ }10100 \qquad \mathrm{(E) \ } 10102$

Solution

We begin to evaluate the first couple of terms of the sequence, hoping to find a pattern: $2, 4, 8, 14, 22, ....$. We notice that the difference between succesive terms of the sequence are $2, 4, 6, 8, ....$, a clear pattern. We can see that this pattern continues infinitely because of the recursive definition: each term is the previous term plus the next even number. Therefore, since the differences of consecutive terms form an arithmetic sequence, then the terms satisfy a quadratic, specifically, the one that contains the points $(1, 2), (2, 4),$, and $(3, 8)$. Let the quadratic be $f(x)=ax^2+bx+c$, so:

$a+b+c=2$ (1)

$4a+2b+c=4$ (2)

$9a+3b+c=8$ (3)

Subtracting (1) from (2) and (2) from (3) yields the two-variable system of equations

$3a+b=2$ (4)

$5a+b=4$ (5)

We can subtract (4) from (5) to find that $2a=2$, so $a=1$. Substituting this back in yields $b=-1$, and substituting these back into one of the original equations yields $c=2$, so the closed form for the terms is

$f(x)=x^2-x+2$, or

$a_n=n^2-n+2$.

Substituting in $n=100$ yields $a_{100}=100^2-100+2=9902, \boxed{\text{B}}$.

Alternate Solution

Term $a_{100}$ can be viewed as the first term, $2$, plus the arithmetic series $2 + 4 +\cdots + 198$.

By summing the numbers that form the arithmetic sequence with $a_{1}$, we get $2 + \dfrac{2 + 198}{2}\cdot99 = 9902$.

See Also

1984 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 13
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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