# 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 $2 + \dfrac{2 + 198}{2}\cdot99 = 9902$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 