Difference between revisions of "1969 AHSME Problems/Problem 32"
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The first three terms of the sequence are <math>5</math>, <math>12</math>, and <math>23</math>. From there, a [[system of equations]] can be written. | The first three terms of the sequence are <math>5</math>, <math>12</math>, and <math>23</math>. From there, a [[system of equations]] can be written. | ||
<cmath>a+b+c=5</cmath> | <cmath>a+b+c=5</cmath> | ||
− | <cmath>4a+2b+c= | + | <cmath>4a+2b+c=8</cmath> |
− | <cmath>9a+3b+c= | + | <cmath>9a+3b+c=15</cmath> |
− | Solve the system to get <math>a=2</math>, <math>b= | + | Solve the system to get <math>a=2</math>, <math>b=-3</math>, and <math>c=6</math>. The sum of the coefficients is <math>\boxed{\textbf{(C) } 5}</math>. |
− | Note: | + | Note: Solving the system is extra work, as the answer is described by the first equation. The sum of the coefficients (<math>a + b + c</math>) is just 5 by the first equation. |
== See also == | == See also == |
Revision as of 03:19, 14 July 2019
Problem
Let a sequence be defined by and the relationship If is expressed as a polynomial in , the algebraic sum of its coefficients is:
Solution
Note that the first differences create a linear function, so the sequence is quadratic.
The first three terms of the sequence are , , and . From there, a system of equations can be written. Solve the system to get , , and . The sum of the coefficients is .
Note: Solving the system is extra work, as the answer is described by the first equation. The sum of the coefficients () is just 5 by the first equation.
See also
1969 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 31 |
Followed by Problem 33 | |
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 • 31 • 32 • 33 • 34 • 35 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.