1969 AHSME Problems/Problem 32

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 () is just 5 by the first equation.

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
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1969 AHSME Problems/Problem 32

Problem

Solution

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