easy question

Let S(n) be the statement: 1 x3 +2x3 ^2 + 3x3 ^3 + ... + n x 3 ^n = 3/4 [1+ (2n - 1 ) 3 ^n]
When n = 1, LHS = 1x3 = 3
RHS = (3/4)(1+3) = 3/4 x 4 = 3 = LHS
S(1) is true.
Assume S(k) is true, i.e. 1 x3 +2x3 ^2 + 3x3 ^3 + ... + k x 3 ^k = (3/4)[1+(2k-1) 3^k]
When n = k+1,
LHS = 1 x3 +2x3 ^2 + 3x3 ^3 + ... + k x 3 ^k + (k+1)x3^(k+1)
= (3/4)[1+ (2k - 1 ) 3^k]+ (k+1)x3^(k+1)
= (3/4)[ 1+(2k - 1 ) 3^k + (4/3) (k+1)3^(k+1)]
= (3/4)[ 1+(2k - 1 ) 3^k + 4(k+1)3^k]
= (3/4)[ 1+(2k -1+4k+4 ) 3^k ]
= (3/4)[1+3(2k+1)3^k]
= (3/4)[1 + [2(k+1)-1]3^(k+1)]
S(k+1) is true.
By M.I., S(n) is true for all positive integers n.