{[f(1)=0],[f(n)=f(n-1)+2]:} Find an explicit formula for f(n). f(n)=

Given Base Case and Recursive Formula: We are given the base case f(1) = 0 and the recursive formula f(n) = f(n-1) + 2. To find an explicit formula, we need to see how the function behaves as n increases.Calculate First Few Values: Let's calculate the first few values of the function to identify a pattern: f(1) = 0 (given) f(2) = f(1) + 2 = 0 + 2 = 2 f(3) = f(2) + 2 = 2 + 2 = 4 f(4) = f(3) + 2 = 4 + 2 = 6 We can see that each time n increases by 1, the function value increases by 2.Identify Pattern: From the pattern, we can infer that f(n) increases by 2 for each increment in n starting from f(1). So, we can express f(n) as 2 times the number of increments from 1 to n. Since f(1) = 0, the number of increments from 1 to n is (n - 1).Infer Increment Relationship: Therefore, the explicit formula for f(n) is f(n) = 2 * (n - 1).Verify Formula: Let's verify the formula with the base case and one other value: For n = 1, f(1) = 2 * (1 - 1) = 2 * 0 = 0, which matches the given base case. For n = 3, f(3) = 2 * (3 - 1) = 2 * 2 = 4, which matches the value we calculated earlier.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

{[f(1)=0],[f(n)=f(n-1)+2]:}
Find an explicit formula for
f(n).

f(n)=

$\left\{\begin{array}{l} f(1)=0 \\ f(n)=f(n-1)+2 \end{array}\right.$ $\newline$ Find an explicit formula for $f(n)$ . $\newline$ $f(n)=$

View step-by-step help

Home

Math Problems

Algebra 2

Write a formula for an arithmetic sequence

Full solution

\left\{\begin{array}{l} f(1)=0 \\ f(n)=f(n-1)+2 \end{array}\right.

\newline

Find an explicit formula for

f(n)

\newline

f(n)=

Given Base Case and Recursive Formula: We are given the base case $f(1) = 0$ and the recursive formula $f(n) = f(n-1) + 2$ . To find an explicit formula, we need to see how the function behaves as $n$ increases.
Calculate First Few Values: Let's calculate the first few values of the function to identify a pattern: $\newline$ $f(1) = 0$ (given) $\newline$ $f(2) = f(1) + 2 = 0 + 2 = 2$ $\newline$ $f(3) = f(2) + 2 = 2 + 2 = 4$ $\newline$ $f(4) = f(3) + 2 = 4 + 2 = 6$ $\newline$ We can see that each time $n$ increases by $1$ , the function value increases by $2$ .
Identify Pattern: From the pattern, we can infer that $f(n)$ increases by $2$ for each increment in $n$ starting from $f(1)$ . So, we can express $f(n)$ as $2$ times the number of increments from $1$ to $n$ . Since $f(1) = 0$ , the number of increments from $1$ to $n$ is $2$ $1$ .
Infer Increment Relationship: Therefore, the explicit formula for $f(n)$ is $f(n) = 2 \times (n - 1)$ .
Verify Formula: Let's verify the formula with the base case and one other value: $\newline$ For $n = 1$ , $f(1) = 2 \times (1 - 1) = 2 \times 0 = 0$ , which matches the given base case. $\newline$ For $n = 3$ , $f(3) = 2 \times (3 - 1) = 2 \times 2 = 4$ , which matches the value we calculated earlier.