1976 AHSME Problems/Problem 25

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

1976 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 26
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1976 AHSME Problems/Problem 25

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