Difference between revisions of "1992 AIME Problems/Problem 8"

Revision as of 22:39, 27 February 2008

Problem

For any sequence of real numbers $A=(a_1,a_2,a_3,\ldots)$ , define $\Delta A^{}_{}$ to be the sequence $(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)$ , whose $n^\mbox{th}_{}$ (Error compiling LaTeX. Unknown error_msg) term is $a_{n+1}-a_n^{}$ . Suppose that all of the terms of the sequence $\Delta(\Delta A^{}_{})$ are $1^{}_{}$ , and that $a_{19}=a_{92}^{}=0$ . Find $a_1^{}$ .

Solution

Since the second differences are all $1$ and $a_{19}=a_{92}^{}=0$ , $a_n$ can be expressed explicitly by the quadratic: $a_n=\frac{1}{2!}(n-19)(n-92)$ .

Thus, $a_1=\frac{1}{2!}(1-19)(1-92)=\boxed{819}$ .

@@ Line 3: / Line 3: @@
 == Solution ==
-Since the second differences of the sequence is constant, the sequence can be expressed explicitly as a quadratic in terms of <math>n</math>.
+Since the second differences are all <math>1</math> and <math>a_{19}=a_{92}^{}=0</math>, <math>a_n</math> can be expressed explicitly by the quadratic: <math>a_n=\frac{1}{2!}(n-19)(n-92)</math>.
-Since <math>a_{19}=a_{92}^{}=0</math>, and the second differences are all <math>1</math>, <math>a_n=\frac{1}{2!}(n-19)(n-92)</math>.
 Thus, <math>a_1=\frac{1}{2!}(1-19)(1-92)=\boxed{819}</math>.

1992 AIME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

Difference between revisions of "1992 AIME Problems/Problem 8"

Revision as of 22:39, 27 February 2008

Problem

Solution

See also