Find the value of "n" where 3^48 + 3^1996 +3^3943+3^3n.

• 3^48 + 3^1996 + 3^3943 + 3^3n
  = (3^16)^3 + 3*(3^16)^2*3^n + 3*3^16*(3^n)^2+ (3^n)^3 -----(1)
  
  also (1) has the form of (a+b)^3 =
  a^3 + 3*a^2*b + 3*a*b^2 + b^3 -------(2)
  
  from (1) & (2)
  => 3*(3^16)^2*3^n=3^1996
  => 3^(33+n)=3^1996
  => (33+n)=1996
  => n=1963
  
• 10 years agoHelpfull: Yes(27) No(2)
• (3^16 ) ^3+3^1996+3^3943+(3^n)^3
  So it is equivalent to a^3+3a^2b+3ab^2+b^3
  So,3*(3^16)^2*3^n=3^1996
  3^(33+n)=3^1996
  n=1963
• 10 years agoHelpfull: Yes(4) No(0)
• Is this question complete??
• 10 years agoHelpfull: Yes(1) No(0)
• dif is 1948 thn 1947 so third must be 1946.
  now 3^(3943+1946)=3^(5889)=3^(3*1963)
  so n=1963
• 10 years agoHelpfull: Yes(1) No(1)
• @vikram no the question is not incomplete
  ans is 1963 as all solved above..
• 10 years agoHelpfull: Yes(0) No(0)
• @rakesh : bingo★
• 10 years agoHelpfull: Yes(0) No(0)