9^n+9^n-1+........+9^3+9^2+9^1. if n is greater than and equal to 4 (i.e 4,5,6,....) then for all values of n series is always divisible by

a. 7380
b. 729
c. 90
d. 9

• as it is a GP, with a =9^n and r=1/9 the sum would be , (9^n(1-(1/9)^n))/(1-1/9). on solving we get ,
  
  9^n/(8/9) = 9^(n+1)/8
  
  now when you put n=4 you will notice that the result will always be divisible by 9^5 and lower but as the options have 9^3=729 and 9^1=9 you could go with either of them.
• 7 years agoHelpfull: Yes(5) No(2)
• taking n=4,
  9^4+9^3+9^2+9^1=9(9^3+9^2+9^1+1) , which is divisible by 9
  
  taking n=5 we have 9^5+9^4+9^3+9^2+9^1=9(9^4+9^3+9^2+9^1+1)
  which is again divisible by 9
  
  so, the answer is d.
• 7 years agoHelpfull: Yes(3) No(3)
• taking in reverse
  9+9^2+9^3+9^4
  which is a gp
  so sum is 9(9^3-1)/(9-1)=91*9 which is only divisible by 9.
• 7 years agoHelpfull: Yes(0) No(0)
• Yes, 9 is the right answer.
• 7 years agoHelpfull: Yes(0) No(0)
• d
  let n=4
  9(9^3+9^2+9^1+9^0) is always divisible by 9
• 5 years agoHelpfull: Yes(0) No(0)