If the sequence {a_n} is defined by a_1&=2, a_n+1&=a_n+2n (n≥1), then a_100 equals

1984 AMC 12 Problem 12

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12.

If the sequence

\{a_n\}

is defined by

\begin{aligned} a_1&=2,\\ a_{n+1}&=a_n+2n\quad(n\ge1), \end{aligned}

then

a_{100}

equals

9900

9902

9904

10100

10102

Add all increments from

a_1

through

a_{100}

Use

a_{100}=2+2(1+2+\cdots+99)

Telescoping the recurrence gives

\begin{aligned} a_{100} &=a_1+2\sum_{n=1}^{99}n\\ &=2+2\cdot\frac{99\cdot100}{2}\\ &=9902. \end{aligned}

Therefore, the correct answer is B.