Difference between revisions of "1969 AHSME Problems/Problem 32"

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== Solution ==
 
== Solution ==
<math>\fbox{C}</math>
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Note that the first differences create a linear function, so the sequence <math>{u_n}</math> is quadratic.
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The first three terms of the sequence are <math>5</math>, <math>8</math>, and <math>15</math>.  From there, a [[system of equations]] can be written.
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<cmath>a+b+c=5</cmath>
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<cmath>4a+2b+c=8</cmath>
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<cmath>9a+3b+c=15</cmath>
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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>.
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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 ==

Latest revision as of 03:19, 14 July 2019

Problem

Let a sequence $\{u_n\}$ be defined by $u_1=5$ and the relationship $u_{n+1}-u_n=3+4(n-1), n=1,2,3\cdots.$If $u_n$ is expressed as a polynomial in $n$, the algebraic sum of its coefficients is:

$\text{(A) 3} \quad \text{(B) 4} \quad \text{(C) 5} \quad \text{(D) 6} \quad \text{(E) 11}$

Solution

Note that the first differences create a linear function, so the sequence ${u_n}$ is quadratic.

The first three terms of the sequence are $5$, $8$, and $15$. From there, a system of equations can be written. \[a+b+c=5\] \[4a+2b+c=8\] \[9a+3b+c=15\] Solve the system to get $a=2$, $b=-3$, and $c=6$. The sum of the coefficients is $\boxed{\textbf{(C) } 5}$.

Note: Solving the system is extra work, as the answer is described by the first equation. The sum of the coefficients ($a + b + c$) is just 5 by the first equation.

See also

1969 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 31
Followed by
Problem 33
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