Prove the following by using the principle of mathematical induction for all `n in N`:`1. 2 + 2. 3 + 3. 4 + ... + n(n + 1)=[(n(n+1)(n+2))/3]`

Prove the following by using the principle of mathematical induction f

1.	Prove the following by using the principle of mathematical induction for all `n in N`:`1. 2 + 2. 3 + 3. 4 + ... + n(n + 1)=[(n(n+1)(n+2))/3]`
Answer» Let the given statement be P(n). Then, `P(n): 12+23+34+..+n(n+1)= 1/3 n(n+1)(n+2)`. When n = 1, LHS = ` 12=2 and RHS = 1/3 xx 1 xx 2 xx(1+2)=2`. `:. ` 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) : 12+23+34+...+k(k+1) = 1/3 k(k+1)(k+2)`. ....(i) Now, `12+23+34+...+k(k+1)+(k+1)(k+2)` ` = {12+23+34+...+k(k+1)+(k+1)(k+2)` ` = 1/3 k(k+1)(k+2)+(k+1)(k+2)" "` [using (i)] ` = 1/3 [k(k+1)(k+2)+3(k+1)(k+2)]= 1/3 (k+1)(k+2)(k+3)`. ` :. " " P(k+1) : 1* 2 + 2 + 2 * 3 + 3 * 4 +...+(k+1)(k+2)=1/3 (k+1)(k+2)(k+3)`. This shows that P(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 ` 12+23+3*4+...+n(n+1) = 1/3 n(n+1) (n+2)" for all " n in N`.