Linear combinations of two GPs

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We explore a classic link between sequences: linear combinations of two geometric progressions (GPs) produce terms that satisfy a second-order homogeneous recurrence. This connection sits squarely within the sequences and series topic of the HSC Extension 1 syllabus and provides a direct path to solving recurrence equations—by expressing unknown sequences as combinations of GPs with the characteristic roots. Working through the problem will sharpen your ability to move between closed-form expressions and recurrences, a skill that underpins many advanced problems in Extension 2 as well.

Problem Statement

Let Tn=arn1T_n=ar^{n-1} and Un=ARn1U_n=AR^{n-1}. Define Wn=Tn+UnW_n=T_n+U_n.

  • Show (Wn)(W_n) satisfies
Wn+2(r+R)Wn+1+rRWn=0.W_{n+2}-(r+R)W_{n+1}+rRW_n=0.
  • Solve
Xn+25Xn+1+6Xn=0,X1=5, X2=13.X_{n+2}-5X_{n+1}+6X_n=0,\quad X_1=5,\ X_2=13.

Hints

For (ii), the characteristic roots are 22 and 33.


Solutions

For (i), we simply substitute the definition of WnW_n into the left-hand side and see that it collapses to zero. Because WnW_n is the sum of two GPs with common ratios rr and RR, the recurrence relation naturally emerges:

Substitute Wn=arn1+ARn1W_n=ar^{n-1}+AR^{n-1} directly to verify (i).

For (ii), we recognise that the recurrence matches the form from part (i) with r=2r=2 and R=3R=3 (since 5=2+35=2+3 and 6=2×36=2\times3). Therefore its solution is a linear combination of the two GPs with those common ratios. We let

Xn=α2n1+β3n1.X_n=\alpha 2^{n-1}+\beta 3^{n-1}.

Using the given initial terms X1=5X_1=5 and X2=13X_2=13 gives the system

α+β=5,2α+3β=13\alpha+\beta=5,\quad 2\alpha+3\beta=13

gives α=2, β=3\alpha=2,\ \beta=3. Hence

Xn=22n1+33n1=2n+3n.X_n=2\cdot2^{n-1}+3\cdot3^{n-1}=2^n+3^n.

Takeaways

  • A sum of two geometric progressions with distinct ratios rr and RR satisfies Wn+2(r+R)Wn+1+rRWn=0W_{n+2}-(r+R)W_{n+1}+rRW_n=0, a second-order linear recurrence whose characteristic equation has roots rr and RR.
  • Conversely, a recurrence of the form Xn+2(r+R)Xn+1+rRXn=0X_{n+2}-(r+R)X_{n+1}+rRX_n=0 can be solved by writing Xn=αrn1+βRn1X_n=\alpha r^{n-1}+\beta R^{n-1} (provided rRr\neq R); the initial values determine α\alpha and β\beta.
  • This technique—expressing a sequence as a linear combination of geometric progressions—is a powerful algebraic shortcut that avoids more cumbersome iterative methods.

Further Readings

HSC Mechanics, HSC Integrals, HSC Induction, HSC Functions

Written by Vu Hung Nguyen

Mathematics Educator · LinkedIn · About