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lesson summary heres an arithmetic sequence f: 6, 10, 14, 18, 22, .... …

Question

lesson summary heres an arithmetic sequence f: 6, 10, 14, 18, 22, .... in this sequence, each term is 4 more than the previous term. one recursive definition of this sequence is f(1)=6, for f(n)=f(n - 1)+4 for n≥2. we could also write f(0)=6, f(n)=f(n - 1)+4, for n≥1 since it generates the same sequence. neither of these definitions is better than the other, we just have to remember how we chose to define the \first term\ of the sequence: f(1) or f(0). lets use f(1) for now. while defining a sequence recursively works to calculate the current term from the previous term, if we wanted to calculate, say, f(100), it would mean calculating all the terms up to f(99) to get there! lets think of a better way. since we know that each term has an increasing number of fours, we could write the terms organized in a table like the one shown here. n f(n) 1 6 + 0=6 + 4(0)=6 2 6 + 4=6 + 4(1)=10 3 6 + 4 + 4=6 + 4(2)=14 4 6 + 4 + 4 + 4=6 + 4(3)=18 5 6 + 4 + 4 + 4 + 4=6 + 4(4)=22

Explanation:

Step1: Identify arithmetic - sequence formula

The general formula for an arithmetic sequence is $f(n)=f(1)+(n - 1)d$, where $f(1)$ is the first - term and $d$ is the common difference.

Step2: Determine values of $f(1)$ and $d$

From the sequence $6,10,14,18,22,\cdots$, we have $f(1)=6$ and $d = 4$.

Step3: Substitute values into the formula

Substituting $f(1)=6$ and $d = 4$ into the formula $f(n)=f(1)+(n - 1)d$, we get $f(n)=6+(n - 1)\times4$.

Step4: Simplify the formula

Expand the expression: $f(n)=6 + 4n-4=4n + 2$.

Answer:

The explicit formula for the given arithmetic sequence is $f(n)=4n + 2$.