Let , , , and for . Find , , , , and .

Understand the Question We are given the first three terms of a sequence: , , and . For any term from , the recurrence relation is , meaning each new term is the sum of the preceding three terms (a Tribonacci-like sequence). We calculate and step-by-step in sequence, using the newly computed values for subsequent terms. Step 1 · Find Given , , , and . For : Step 2 · Find For : Step 3 · Find For : Step 4 · Find For : Step 5 · Find For : Answer > Common Mistakes Adding Only Two Terms: Confusing this 3-term recurrence relation with the standard Fibonacci sequence () and forgetting to add the third preceding term (). Cascading Arithmetic Errors: An addition error in an earlier term (such as or ) will carry forward and cause all subsequent terms to be incorrect.

Predicting What Comes Next: Sequences and Progressions | Exercise 8.1

Question 6

Let $T_1 = 1$ , $T_2 = 2$ , $T_3 = 4$ , and $T_n = T_{n-1} + T_{n-2} + T_{n-3}$ for $n \ge 4$ . Find $T_4$ , $T_5$ , $T_6$ , $T_7$ , and $T_8$ .

Solution

Understand the Question

We are given the first three terms of a sequence: $T_1 = 1$ , $T_2 = 2$ , and $T_3 = 4$ .
For any term from $n \ge 4$ , the recurrence relation is $T_n = T_{n-1} + T_{n-2} + T_{n-3}$ , meaning each new term is the sum of the preceding three terms (a Tribonacci-like sequence).
We calculate $T_4, T_5, T_6, T_7,$ and $T_8$ step-by-step in sequence, using the newly computed values for subsequent terms.

Step 1 · Find $T_4$

Given $T_1 = 1$ , $T_2 = 2$ , $T_3 = 4$ , and $T_n = T_{n-1} + T_{n-2} + T_{n-3}$ . Diagram 1

For $n = 4$ :

\begin{aligned} T_4 &= T_3 + T_2 + T_1 \\[0.6em] &= 4 + 2 + 1 \\[0.6em] &= 7 \end{aligned}

Step 2 · Find $T_5$

For $n = 5$ :

\begin{aligned} T_5 &= T_4 + T_3 + T_2 \\[0.6em] &= 7 + 4 + 2 \\[0.6em] &= 13 \end{aligned}

Step 3 · Find $T_6$

For $n = 6$ :

\begin{aligned} T_6 &= T_5 + T_4 + T_3 \\[0.6em] &= 13 + 7 + 4 \\[0.6em] &= 24 \end{aligned}

Step 4 · Find $T_7$

For $n = 7$ :

\begin{aligned} T_7 &= T_6 + T_5 + T_4 \\[0.6em] &= 24 + 13 + 7 \\[0.6em] &= 44 \end{aligned}

Step 5 · Find $T_8$

For $n = 8$ :

\begin{aligned} T_8 &= T_7 + T_6 + T_5 \\[0.6em] &= 44 + 24 + 13 \\[0.6em] &= 81 \end{aligned}

Answer

$T_4 = 7, \quad T_5 = 13, \quad T_6 = 24, \quad T_7 = 44, \quad T_8 = 81$

Common Mistakes

Adding Only Two Terms: Confusing this 3-term recurrence relation with the standard Fibonacci sequence ( $T_n = T_{n-1} + T_{n-2}$ ) and forgetting to add the third preceding term ( $T_{n-3}$ ).
Cascading Arithmetic Errors: An addition error in an earlier term (such as $T_4$ or $T_5$ ) will carry forward and cause all subsequent terms to be incorrect.

Question 6

More questions in Exercise 8.1