+0

# function

0
44
1

f satisfies

f(x + y) = f(x) + xy + f(y)

for all x, y.  Also, f(5) = 70.  Find f(9).

May 8, 2020

#1
+24992
+1

function

$$f(x + y) = f(x) + xy + f(y)$$ for all $$x,~ y$$

Also, $$f(5) = 70$$

Find $$f(9)$$.

$$\begin{array}{|rcll|} \hline f(1 + 1) = f(2) &=& f(1) + 1\cdot 1 + f(1) \\ \mathbf{ f(2) } &=& \mathbf{ 1+2f(1) } \\\\ f(1 + 2) = f(3) &=& f(1) + 1\cdot 2 + f(2) \\ f(1 + 1) = f(2) &=& f(1) + 1\cdot 1 + f(1) \\ f(3) &=& 2+f(1)+f(2) \quad | \quad \mathbf{ f(2) =1+2f(1) } \\ f(3) &=& 2+f(1)+ 1+2f(1) \\ \mathbf{ f(3) } &=& \mathbf{ 3+3f(1) } \\\\ f(1 + 3) = f(4) &=& f(1) + 1\cdot 3 + f(3) \\ f(4) &=& 3+f(1)+f(3) \quad | \quad \mathbf{ f(3) =3+3f(1) } \\ f(4) &=& 3+f(1)+ 3+3f(1) \\ \mathbf{ f(4) } &=& \mathbf{ 6+4f(1) } \\\\ f(1 + 4) = f(5) &=& f(1) + 1\cdot 4 + f(4) \\ f(5) &=& 4+f(1)+f(4) \quad | \quad \mathbf{ f(4) =6+4f(1) } \\ f(5) &=& 4+f(1)+ 6+4f(1) \\ \mathbf{ f(5) } &=& \mathbf{ 10+5f(1) } \quad | \quad \boxed{f(5) = 70} \\\\ 70 &=& 10+5f(1) \\ 60 &=& 5f(1) \\ \mathbf{f(1)} &=& \mathbf{12} \\\\ f(4 + 5) = f(9) &=& f(4) + 4\cdot 5 + f(5) \\ f(9) &=& 20+ f(4) + f(5) \quad | \quad \mathbf{ f(4) =6+4f(1) } \\ f(9) &=& 20+ 6+4f(1) + f(5) \\ f(9) &=& 26+4f(1) + f(5) \quad | \quad \mathbf{ f(1) =12 },\ \mathbf{ f(5) =70 } \\ f(9) &=& 26+4\cdot 12 + 70 \\ f(9) &=& 26+48+ 70 \\ \mathbf{f(9)} &=& \mathbf{144} \\ \hline \end{array}$$

May 11, 2020