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

(Solution to Problem 32)
(Solution)
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<cmath>9a+3b+c=23</cmath>
 
<cmath>9a+3b+c=23</cmath>
 
Solve the system to get <math>a=2</math>, <math>b=1</math>, and <math>c=2</math>.  The sum of the coefficients is <math>\boxed{\textbf{(C) } 5}</math>.
 
Solve the system to get <math>a=2</math>, <math>b=1</math>, and <math>c=2</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.
  
 
== See also ==
 
== See also ==

Revision as of 14:24, 21 November 2018

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 $5$, $12$, and $23$. From there, a system of equations can be written. \[a+b+c=5\] \[4a+2b+c=12\] \[9a+3b+c=23\] Solve the system to get $a=2$, $b=1$, and $c=2$. 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.

See also

1969 AHSC (ProblemsAnswer KeyResources)
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
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