Using the principle of mathematical induction prove that `:``1. 3+2. 3^2+3. 3^3++n .3^n=((2n+1)3^(n+1)+3)/4^`for all `n in N`.

Using the principle of mathematical induction prove that `:``1. 3+2. 3

1.	Using the principle of mathematical induction prove that `:``1. 3+2. 3^2+3. 3^3++n .3^n=((2n+1)3^(n+1)+3)/4^`for all `n in N`.
Answer» Let the given statement be P(n) . Then, `P(n) : 13+23^(2)+33^(3)+…+n3^(n) = ((2n-1)3^(n+1)+3)/4` . Putting n = 1 in the given statement, we get LHS ` = 13 and RHS = ((2XX1-1)3^(1+1)+3)/4=(1xx9+3)/4 = 12/4 = 3`. `:. ` LHS = RHS. Thus , P(1) is true. Let P(k) be true. Then, ` P(k) : 13+23^(2)+33^(3)+...+k3^(k) = ((2k-1)3^(k+1)+3)/4 `. ...(i) ` :. 13+23^(2)+33^(30+...+k3^(k)+(k+1) 3^((k+1))` `=((2k-1)3^(k+1)+3)/4 +(k+1) 3^(k+1)" " ` [using (i)] `=((2k-1)3^(k+1)+3+4(k+1)3^(k+1))/4 = ((2k-1+4k+4)3^(k+1)+3)/4 ` ` = ((6K+3) 3^(k+1)+3)/4 = ({2(k+1)-1}3^((k+1)+1)+3)/4 ` . ` :. P(k+1): 13+23^(2)+33^(3)+...+(k+1)3^(k+1)=({2(k+1)-1}3^((k+1)+1)+3)/4`. This shows that P(k+1) is true, whenever P(k) is true. Thus, P(1) is true and P(k+1) is true, whenever P(k) is true. Hence, by principle of mathematical induction, we have ` 13+23^(2)+33^(3)+...+ n*3^(n) = ((2n-1) 3^(n+1)+3)/4 " for all values of " n in N`.