mathematical induction問題

Let S(n) be the statement 3 + 3^2 + 3^3 + ．．．+ 3^n = 3/2(3^n - 1 )
When n=1,
L.H.S.=3
R.H.S.=3(3^1-1)/2=3
Therefore S(1) is true.
Let S(k) is true.
When n=k+1,
L.H.S.=3(3^k-1)/2+3^(k+1)
=[3^(k+1)-3]/2+3^(k+1)
={3^(k+1)+2[3^(k+1)]-3}/2
=[3^(k+2)-3]/2
=3[3^(k+1)-1]/2
R.H.S.=3[3^(k+1)-1]/2
Therefore by the principle of M.I, S(n) is true for all natural number n.

3 + 32 + 33 + ．．．+ 3n = 3(3n - 1 ) /2
let n=1
LHS = 3
RHS = 3/2(3n - 1 ) = 3(31 - 1 )/2 = 3x2 / 2 = 3
LHS = RHS
assume n=k is true
3 + 32 + 33 + ．．．+ 3k = 3(3k - 1 ) /2
let n = k + 1
LHS = 3 + 32 + 33 + ．．．+ 3k + 3k+1
= 3(3k - 1 ) /2 + 3k+1
= 3(3k - 1 ) /2 + 3x3k
= 3(3k – 1 + 2x3k+1 ) /2
= 3(3x3k – 1) /2
= 3(3k+1 – 1) /2
= RHS
such that the following proposition is true for all positive integers n.