Find the value of “n” where 348 + 31996 + 33943 + 33n.
Option
1) 1963
2) 1964
3) 1960
4) 1991

• 3^48 + 3^1996 + 3^3943+3^3n
  (3^16 )^3 + 3^1996 + 3^3943 + (3^n)^3
  its expansion of (a+b)^3=a^3+3a^2b+3ab^2+b^3
  
  => 3*(3^16)^2*3^n=3^1996
  => 3^(33+n)=3^1996
  => n = 1963
• 10 years agoHelpfull: Yes(36) No(5)
• 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
• 7 years agoHelpfull: Yes(5) No(0)
• plz explain
  
• 8 years agoHelpfull: Yes(0) No(1)