the infinite sum 1+(4/7)+(9/7^2)+(16/7^3)+.........?
options
A>27/14
B>29/13
C>49/27
D>256/147

• S = 1 + 4/7 + 9/7^2 + 16/7^3 + 25/7^4 +...
  Dividing both the side by 7
  S/7 = 1/7 + 4/7^2 + 9/7^3 +...
  Subtracting eqn.
  6S/7 = 1 + 3/7 + 5/7^2 + 7/7^3 + ....
  6S/7^2 = 1/7 + 3/7^2 + 5/7^3
  Subtracting again
  36S/49 = 1 + 2/7 + 2/7^2 + 2/7^3
  take 2/7 outside.....then it will be a onfinite GP.....1 + 1/7 + 1/7^2.....ETC.........ie 7/6.......and u will get 49/27
• 10 years agoHelpfull: Yes(23) No(0)
• @ pranoj plz tm sequence series prh lo...
  
  ans is 49/27. tn = n^2/7^n-1.
• 10 years agoHelpfull: Yes(3) No(0)
• ANSWER--> C
  Because
  1+4/7+9/49+16/343+25/2401+..... = 1.76+...
  now the 6th term = 36/16807= 0.002
  so going through the option we find that 49/27=1.814 which is closest value
  hence it it the right choice
• 10 years agoHelpfull: Yes(0) No(8)