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Let $f: \mathbf{N} \times \mathbf{N} \rightarrow \mathbf{N}$ be a function such that $f((1,1))=2$ and $f((m+1, n))=f((m, n))$ $+2(m+n)$ and $f((m, n+1))=f((m, n))+2$ $(m+n-1), \forall m, n \in \mathbf{N}$, then find $f(2,2)$
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Verified Answer
The correct answer is:
10
Given function $f: \mathbf{N} \times \mathbf{N} \longrightarrow \mathbf{N}$, such that
$$
f(m+1, n)=f(m, n)+2(m+n)
$$
and $f(m, n+1)=f(m, n)+2(m+n-1)$, where $f(1,1)=2$.
$$
\begin{aligned}
\therefore \quad f(2,2) & =f(1+1,2)=f(1,2)+2(1+2) \\
& =f(1,1+1)+6 \\
& =f(1,1)+2(1+1-1)+6 \\
& =f(1,1)+8=2+8=10
\end{aligned}
$$
$$
f(m+1, n)=f(m, n)+2(m+n)
$$
and $f(m, n+1)=f(m, n)+2(m+n-1)$, where $f(1,1)=2$.
$$
\begin{aligned}
\therefore \quad f(2,2) & =f(1+1,2)=f(1,2)+2(1+2) \\
& =f(1,1+1)+6 \\
& =f(1,1)+2(1+1-1)+6 \\
& =f(1,1)+8=2+8=10
\end{aligned}
$$
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