`1.3+2.3^(2)+3.3^(3)+. . .+n.3^(3)=((2n-1)3^(n+1)+3)/(4)`

Explanation

`P(n) : 1 xx 3 xx + 2 xx 3^(2) + 3 xx 3^(3) + "…." n xx 3^(n)`
`= ((2n-1) 3^(n+1) + 3)/(4)`
For `n = 1`
and `L.H.S. = 1 xx 3 = 3`
and `R.H.S. = ((2 xx 1 - 1)3^(1+1)+3)/(4) = (3^(2)+3)/(4) = (12)/(4) = 3`
Thus `P(1)` is true.
Let `P(n)` be true for some `n = k`
i.e, `1 xx 3 + 2 xx 3^(2) + 3 xx 3^(3) + "..... + k xx 3^(k)`
`= ((2k-1)3^(k+1)+3)/(4)`
Now, we have to prove that `P(n)` is true for `n = k 1` ltb rgt i.e, `1 xx 3 + 2 xx 3^(2) + 3 xx 3^(3) + "...." + k +1 xx 3^(k+1)`
`= ((2k+1)3^(k+2)+3)/(4)`
Adding `(k+1) xx 3^(k+1)` both sides of `(1)`, we get
`1 xx 3 + 2 xx 3^(2) + 3 xx 3^(3) + "...." + k xx 3^(4) + (k+1) xx 3^(k+1)`
`= ((2k-1)3^(k+1)+3)/(4)+(k+1)xx3^(k+1)`
`= ((2k-1)3^(k+1)+3+4(k+1)3^(k+1))/(4)`
`= (3^(k+1)[2k-1+4(k+1)]+3)/(4)`
`= (3^(k+1)(6k+3)+3)/(4)`
` = (3^((k+1)+1)(2k+1)+3)/(4)`
`= ((2k+1)3^(k+2)+3)/(4)`
Thus `P (k+1)` is true true whenever `P(k)` is true.
Hence by the principle of mathematical induction, statement `P(n)` is true for all natural numbers.