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If $f: N \rightarrow R$ is defined by $f(1)=-1$ and $f(n+1)=3 f(n)+2$ for $n \geq 1$, then $f$ is
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Verified Answer
The correct answer is:
a constant function
We have, $f: N \rightarrow R$ and $f(1)=-1$
Also, $f(n+1)=3 f(n)+2$ for $n \geq 1$
Now, $\operatorname{put} n=1$, we get
$$
\begin{aligned}
f(2) & =3 f(1)+2 \\
& =3(-1)+2=-1
\end{aligned}
$$
Put $n=2$, we get
$$
f(3)=3 f(2)+2=3(-1)+2=-1
$$
So, $f(n)=-1$ for all values of $n$.
Hence, $f(n)$ is a constant function.
Also, $f(n+1)=3 f(n)+2$ for $n \geq 1$
Now, $\operatorname{put} n=1$, we get
$$
\begin{aligned}
f(2) & =3 f(1)+2 \\
& =3(-1)+2=-1
\end{aligned}
$$
Put $n=2$, we get
$$
f(3)=3 f(2)+2=3(-1)+2=-1
$$
So, $f(n)=-1$ for all values of $n$.
Hence, $f(n)$ is a constant function.
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