Question

3×6+6×9+9×12+_____3n×(3n+3)=3n(n+1)(n+2).............solve it by mathematical induction​

Answer 1

We have to prove,

$P(n):\ 3\cdot 6+6\cdot 9+9\cdot 12+...+3n(3n+3)=3n(n+1)(n+2)$

First we have to check whether P(1) is true or not.

$P(1)\\ \\ \text{LHS}:-\ \ \ 3\cdot 6=18\\ \\ \text{RHS}:-\ \ \ 3\cdot 2\cdot 3=18$

So P(1) is true.

Now we can assume P(k) holds true.

$\textsf{Let\ \ \ 3\cdot 6+6\cdot 9+9\cdot 12+...+3k(3k+3)=3k(k+1)(k+2) }$

Let's consider P(k + 1).

$\begin{aligned}&\textsc{LHS}\\ \\ \Longrightarrow\ \ &3\cdot 6+6\cdot 9+9\cdot 12+...+3(k+1)(3(k+1)+3)\\ \\ \Longrightarrow\ \ &3\cdot 6+6\cdot 9+9\cdot 12+...+3k(3k+3)+3(k+1)(3k+3+3)\\ \\ \Longrightarrow\ \ &3k(k+1)(k+2)+3(k+1)(3k+6)\\ \\ \Longrightarrow\ \ &3k(k+1)(k+2)+9(k+1)(k+2)\\ \\ \Longrightarrow\ \ &3(k+1)(k+2)(k+3)\\ \\ \Longrightarrow\ \ &3(k+1)(k+1+1)(k+1+2)\\ \\ \Longrightarrow\ \ &3n(n+1)(n+2)\\ \\ \Longrightarrow\ \ &\textsc{RHS}\end{aligned}$

Hence Proved!

Answer 2

Step-by-step explanation:

Hey mates your answer is here 3.6 + 6.9 + 9.12 + 3n *(3n+3)  = 3n (n+1)(n+2)

1)  let n = 1

      then    3. 6 =  3*1 (1+1) (1+2) = 18      correct.

2)  let the given expression be true for n = k.

     Then  for  n = k+1,

  3.6 + 6.9+9.12+....+ 3k (3k+3) + (3k+1) (3k+4)

   =  3 k(k+1)(k+2) + (3k+3)(3k+6)

  =   3( k+1) (k+2) [ k + 3]

  This is in the form of  3 n (n+1)(n+2)

So that is proved by Induction.

❤️❤️☺️❤️☺️