Find the value of "n" where 348 + 31996 +33943+33n.

• (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(12) No(7)
• this question is wrong d correct question is 3^48+3^1996+3^3943+3^3n.....solved below by jeeva
• 10 years agoHelpfull: Yes(6) No(1)
• jeeva pls explain properly
• 10 years agoHelpfull: Yes(5) No(0)
• Right question is 3^48+3^1996+3^3943+3^3n
  so check the difference b/w to power 1996-48=1948.
  again check then we will get 3943-1996=1947.
  so here we can see that the difference b/w 1st two power and 2nd two power decreased by 1.
  so we can write for 3rd power i.e.
  3n-3943=1946
  n=1963 answer
• 5 years agoHelpfull: Yes(1) No(0)
• my ans is
  33910
• 8 years agoHelpfull: Yes(0) No(0)
• I request anybody to explain this problem clearly
• 6 years agoHelpfull: Yes(0) No(0)