1969 AHSME Problems/Problem 32

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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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