#SERIES-1


In  a  certain  series, the  nth  term  Tn  equals   4Tn+1  + n-1.  If  T1 = 4,then  find  the  value  of  T200.


Solution-

T1=4
T2 = 4T1 + (2-1) = 4^2 +1
T3 = 4T2 + (3-1) = 4^3 +4+2
T4 = 4T3 + (4-1) = 4^4 +4^2 +2*4 +3
T5 = 4T4 + (5-1) = 4^5 +4^3 +2*4^2 +3*4 +4
....
......
.........
...........
T200 = 4^200 + 4^198 +2*4^197 +3*4^196+......+198*4 +199
       =4^200 + Sn
Now, Sn=4^198 +2*4^197 +3*4^196 +.......+198*4 + 199------(i)
Sn /4 = 4^197 +2*4^196 +3*4^195+..........+198 + 199/4--------(ii)
 Eqn (i)-(ii),
 3Sn/4  = 4^198 +4^197 +4^196+.........+4+1-199/4
  3Sn/4 = (4^199 - 1)/(4-1) - 199/4
 3Sn/4  = (4^199 - 1)/3 - 199/4
      Sn  =(4^200-4)/9 -3*199/9
      Sn  =(4^200 - 4-597)/9
      Sn  = (4^200-601)/9
Substituting this value,we get,

T200 = 4^200 +(4^200 - 601)/9
T200 = (10*4^200 - 601)/9 ANS

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