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{[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 $f(n)$ . $\newline$ $f(n)=$ $\newline$ ______________

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Evaluate functions

Full solution

Q.

\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)=

\newline

______________

Given base case and 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 express $f(n)$ in terms of $n$ without the need to calculate all previous function values.
Calculate first few values: Let's calculate the first few values to see if we can 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( $3$ ) + $2$ = $4$ + $2$ = $6$ $\newline$ It seems that $f(n) = 2 \times (n - 1)$ .
Identify pattern: Let's check if our pattern holds for $n = 1$ : $f(1) = 2 \times (1 - 1) = 2 \times 0 = 0$ , which matches the given base case.
Check pattern for $n=1$ : Now let's check if our pattern holds for $n = 2$ : $f(2) = 2 \times (2 - 1) = 2 \times 1 = 2$ , which matches the value we calculated earlier.
Check pattern for $n=2$ : We can now generalize this pattern to find the explicit formula for $f(n)$ : $f(n) = 2 \times (n - 1)$ .

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