Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1976 AHSME Problems/Problem 25"
(Finished sol.)
 
Line 14: Line 14:
 
Note that
 
Note that
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 +
\Delta^1(u_n) &= \left[(n+1)^3+(n+1)\right]-\left[n^3+n\right] \\
 +
&= \left[n^3+3n^2+4n+2\right]-\left[n^3+n\right] \\
 +
&= 3n^2+3n+2, \\
 +
\Delta^2(u_n) &= \left[3(n+1)^2+3(n+1)+2\right]-\left[3n^2+3n+2\right] \\
 +
&= \left[3n^2+9n+8\right]-\left[3n^2+3n+2\right] \\
 +
&= 6n+6, \\
 +
\Delta^3(u_n) &= \left[6(n+1)+6\right]-\left[6n+6\right] \\
 +
&= \left[6n+12\right]-\left[6n+6\right] \\
 +
&= 6, \\
 +
\Delta^4(u_n) &= 6-6 \\
 +
&= 0.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 
Therefore, the answer is <math>\boxed{\textbf{(D)}}.</math>
 
Therefore, the answer is <math>\boxed{\textbf{(D)}}.</math>

Latest revision as of 11:06, 7 September 2021

Problem

For a sequence $u_1,u_2\dots$, define $\Delta^1(u_n)=u_{n+1}-u_n$ and, for all integer $k>1, \Delta^k(u_n)=\Delta^1(\Delta^{k-1}(u_n))$. If $u_n=n^3+n$, then $\Delta^k(u_n)=0$ for all $n$

$\textbf{(A) }\text{if }k=1\qquad \\ \textbf{(B) }\text{if }k=2,\text{ but not if }k=1\qquad \\ \textbf{(C) }\text{if }k=3,\text{ but not if }k=2\qquad \\ \textbf{(D) }\text{if }k=4,\text{ but not if }k=3\qquad\\ \textbf{(E) }\text{for no value of }k$

Solution

Note that \begin{align*} \Delta^1(u_n) &= \left[(n+1)^3+(n+1)\right]-\left[n^3+n\right] \\ &= \left[n^3+3n^2+4n+2\right]-\left[n^3+n\right] \\ &= 3n^2+3n+2, \\ \Delta^2(u_n) &= \left[3(n+1)^2+3(n+1)+2\right]-\left[3n^2+3n+2\right] \\ &= \left[3n^2+9n+8\right]-\left[3n^2+3n+2\right] \\ &= 6n+6, \\ \Delta^3(u_n) &= \left[6(n+1)+6\right]-\left[6n+6\right] \\ &= \left[6n+12\right]-\left[6n+6\right] \\ &= 6, \\ \Delta^4(u_n) &= 6-6 \\ &= 0. \end{align*} Therefore, the answer is $\boxed{\textbf{(D)}}.$

More generally, we have $\Delta^k(u_n)=0$ for all $n$ if $k\geq4.$

~MRENTHUSIASM

See Also

1976 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Problem 26
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
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1976_AHSME_Problems/Problem_25&oldid=161808"