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

Explanation

Let the given statement be P(n). Then,
`P(n): 1*2*3+2*3*4+..+n(n+1)(n+2)= 1/4 n (n+1)(n+2)(n+3)`.
When n = 1, we have
LHS = ` 1*2*3 = 6 and RHS = 1/4 xx12xx3xx4 = 6`.
` :. ` LHS=RHS.
Thus, the given statement is true for n = 1, i.e., P(1) is true.
Let P(k) be true. Then,
`P(k) : 1*2*3+2*3*4+...+k(k+1)(k+2)`
` = 1/4 k(k+1)(k+2)(k+3)`. ...(i)
Now, ` 1*2*3+2*3*4+..+k(k+1)(k+2)+(k+1)(k+2)(k+3) `
` = {1*2*3+2*3*4+...+k(k+1)(k+2)}+(k+1)(k+2)(k+3)`
` = 1/4 k(k+1)(k+2)(k+3)+(4(k+1)(k+2)(k+3))/4 ` [using (i)]
` 1/4 (k+1)(k+2)(k+3)(k+4)`.
`:. P(k+1): 1 * 2 *3+2*3*4+...+(k+1)(k+2)(k+3)`
` = 1/4 (k+1)(k+2)(k+3)(k+4)`.
Thus, P(k) (k+1) is true, whenever P(k) is true.
P(1) is true and P(k+1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, we have
` 1 *2*3+2*3*4+...+n(n+1)(n+2)= 1/4 n(n+1)(n+2)(n+3) " for all values of " n in N`.