F(1)=1, where f(x+y)=f(x)+f(y)+8xy-2. Find f(7).

• f(2)=f(1+1)=f(1)+f(1)+8*1*1-2=1+1+8-2=8
  f(3)=f(2+1)=f(2)+f(1)+8*2*1-2=8+1+16-2=23
  f(4)=f(3+1)=f(3)+f(1)+8*3*1-2=23+1+24-2=46
  f(7)=f(3+4)=f(3)+f(4)+8*3*4-2=23+46+96-2=163
• 11 years agoHelpfull: Yes(23) No(0)
• f(1)=1
  ans- 163
  f(2)=f(1+1)=f(1)+f(1)+8*1*1-2=8
  f(3)=f(1+2)=23
  f(4)=f(1+3)=46
  so f(7)=f(3+4)=163
• 11 years agoHelpfull: Yes(4) No(0)
• ans: 163
  f(7)=f(1+6)=f(1)+f(6)+8(1)(6)-2 = 47+f(6)
  f(6)=f(1+5)=f(1)+f(5)+8(1)(5)-2 = 39+f(5)
  f(5)=f(1+4)=f(1)+f(4)+8(1)(4)-2 = 31+f(4)
  f(4)=f(1+3)=f(1)+f(3)+8(1)(3)-2 = 23+f(3)
  f(3)=f(1+2)=f(1)+f(2)+8(1)(2)-2 = 15+f(2)
  f(1)=f(1+1)=f(1)+f(1)+8(1)(1)-2 = 8
  So adding all those ans is:163
• 11 years agoHelpfull: Yes(1) No(0)
• 171.
  following the given pattern you can get the ans.
• 11 years agoHelpfull: Yes(0) No(6)
• f(1)=1
  f(1+1)=1+1+8-2=8
  f(2+2)=8+8+32-2=46
  f(2+1)=8+1+16-2=23
  f(4+3)=46+23+128-2=195
• 11 years agoHelpfull: Yes(0) No(4)