Let f: N times N rightarrow N be a function s | vedclass

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(D) Given the function $f: N 	imes N ightarrow N$ with the conditions:$f(1,1) = 2$$f(m+1, n) = f(m, n) + 2(m+n)$$f(m, n+1) = f(m, n) + 2(m+n-1)$To

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$10$

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(D) Given the function $f: N 	imes N ightarrow N$ with the conditions:$f(1,1) = 2$$f(m+1, n) = f(m, n) + 2(m+n)$$f(m, n+1) = f(m, n) + 2(m+n-1)$To find $f(2,2)$,we can use the given recurrence relations:First,calculate $f(2,1)$ using the first relation with $m=1, n=1$:$f(2,1) = f(1,1) + 2(1+1) = 2 + 2(2) = 2 + 4 = 6$Next,calculate $f(2,2)$ using the second relation with $m=2, n=1$:$f(2,2) = f(2,1) + 2(2+1-1) = 6 + 2(2) = 6 + 4 = 10$Alternatively,using the other path:$f(1,2) = f(1,1) + 2(1+1-1) = 2 + 2(1) = 4$$f(2,2) = f(1,2) + 2(1+2) = 4 + 2(3) = 4 + 6 = 10$Thus,$f(2,2) = 10$.

Let $f: N \times N \rightarrow N$ be a function such that $f(1,1)=2$,$f(m+1, n)=f(m, n)+2(m+n)$,and $f(m, n+1)=f(m, n)+2(m+n-1)$ for all $m, n \in N$. Find the value of $f(2,2)$.

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