E2.7 - Quadratic, cubic and exponential sequences (Extended)

Quadratic differences

For $an^2+bn+c$, the constant second difference is $2a$. Therefore $a=\frac{\text{second difference}}{2}$. Subtract $an^2$ from the terms, then find the linear rule $bn+c$ for the remainder.

Worked example 1: a positive linear remainder

Question. Find the nth term of $4,11,22,37,56,\ldots$.

Plan. Find $a$ from the second differences, then remove $an^2$ and fit the remainder.

Thus $a=4/2=2$. Subtract $2n^2$:

The remainder is $n+1$.

Answer. $2n^2+n+1$.

Check. At $n=1,2,3,4,5$, substitution gives $4,11,22,37,56$. The next gap should be $19+4=23$, giving 79; the rule gives $2(6^2)+6+1=79$.

Worked example 2: a negative linear remainder

Question. Find the nth term of $3,6,13,24,39,\ldots$.

Plan. Use the second differences, subtract $2n^2$, then keep the sign of the remainder's gap.

The remainder decreases by 3 and starts at 1. Its rule is $-3n+4$.

Answer. $2n^2-3n+4$.

Check. Position 5 gives $50-15+4=39$. The next gap is 19, giving 58; the rule gives $72-18+4=58$.

Key idea

Differences find the highest-power contribution. Subtraction then reduces the problem to a lower-power rule you already know.

Cubic differences

For $an^3+bn^2+cn+d$, with $a\ne0$, the constant third difference is $6a$. Find $a$ by dividing that difference by 6. Subtract $an^3$, find the quadratic rule for the remainder, then combine the contributions.

Worked example 3: removing a cubic contribution

Question. Find the nth term of $4,15,40,85,156,\ldots$.

Plan. Use third differences to identify the cube part, then subtract it.

The cube coefficient is $6/6=1$. Subtract $n^3=1,8,27,64,125$, leaving $3,7,13,21,31$. This remainder has first differences $4,6,8,10$ and second differences 2, so its square coefficient is 1. Subtract $n^2$ from it, leaving $2,3,4,5,6$, whose rule is $n+1$.

Answer. $n^3+n^2+n+1$.

Check. At positions 1 through 5 the rule gives $4,15,40,85,156$. Continuing the differences gives next second difference 32, next first difference $71+32=103$, and next term 259. The rule gives $216+36+6+1=259$.

Worked example 4: a larger cube coefficient

Question. Find the nth term of $5,23,67,149,281,\ldots$.

Plan. Calculate third differences, remove $2n^3$, and solve the remaining quadratic.

Thus $a=12/6=2$. Subtract $2n^3=2,16,54,128,250$, leaving $3,7,13,21,31$. As before, that remainder is $n^2+n+1$.

Answer. $2n^3+n^2+n+1$.

Check. At $n=4$, $128+16+4+1=149$; at $n=5$, $250+25+5+1=281$. The next differences predict $281+(132+62)=475$, and the rule gives $432+36+6+1=475$.