Let the first term of a series be T₁=6 and its r^ th term T_r=3 T_ r-1 +6^r, r=2,3 , . If the sum of the first n terms of this series is 1 5 (n^2-12 n+39 ) (4 6^n-5 3^n+1 ) , then n is equal to______

Step-by-step solution

aligned & T _ r =3 ~T _ r -1 +6^ r , r =2,3,4, n & T ₂=3 . T ₁+6^2 & ~T ₂=3.6+6^2...(1) aligned aligned & T ₃=3 ~T ₂+6^3 & ~T ₃=3 ~T ₂+6^3 & ~T ₃=3 (3.6+6^2 )+6^3 & ~T ₃=3^2 .6+3.6^2+6^3...(2) aligned aligned & T _ r =3^ r -1 6 [1+ 6 3 + ( 6 3 )^2+ + ( 6 3 )^ r -1 ] & T _ r =3^ r -1 6 (1+2+2^2+ +2^ r -1 ) & T _ r =6 3^ r -1 1 (1-2^ r ) (-1) & T _ r =6 3^ r -1 (2^ r -1 ) & T _ r = 6 3^ r 3 (2^ r -1 ) aligned aligned & T _ r =2 (6^ r -3^ r ) & S _ n =2 (6^ r -3^ r ) & S _ n =2 [ 6 . (6^ n -1 ) 5 - 3 . (3^ n -1 ) 2 ]