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

m (→‎Solution)
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Note that the first differences create a linear function, so the sequence <math>{u_n}</math> is quadratic.
 
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>12</math>, and <math>23</math>.  From there, a [[system of equations]] can be written.
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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.
 
<cmath>a+b+c=5</cmath>
 
<cmath>a+b+c=5</cmath>
 
<cmath>4a+2b+c=8</cmath>
 
<cmath>4a+2b+c=8</cmath>

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 (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

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