Using the principle ofmathematical induction prove that`1/(1. 2. 3)+1/(2. 3. 4)+1/(3. 4. 5)++1/(n(n+1)(n+2))=(n(n+3))/(4(n+1)(n+2)`for all `n in N`

Using the principle ofmathematical induction prove that`1/(1. 2. 3)+1/

1.	Using the principle ofmathematical induction prove that`1/(1. 2. 3)+1/(2. 3. 4)+1/(3. 4. 5)++1/(n(n+1)(n+2))=(n(n+3))/(4(n+1)(n+2)`for all `n in N`
Answer» Let the given statement be P(n). Then, `P(n): 1/(123)+1/(234)+…+1/(n(n+1)(n+2)) = (n(n+3))/(4(n+1)(n+2)).` Putting n = 1 in the given statement , we get LHS = ` 1/(123) = 1/6 and RHS = (1xx(1+3))/(4xx(1+1)(1+2)) = (1xx4)/(4xx2xx3) = 1/6 `. `:." " ` 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/(123)+1/(234)+...+1/(k(k+1)(k+2))=(k(k+3))/(4(k+1)(k+2)).`...(i) Now, `1/(123)+1/(234)+...+1/(k(k+1)(k+2))+1/((k+1)(k+2)(k+3)) ` `={1/(123)+1/(234)+...+1/(k(k+1)(k=2))}+1/((k+1)(k+2)(k+3)) ` `={(k(k+3))/(4(k+1)(k+2))+1/((k+1)(k+2)(k+3))}" "`[using (i)] `=(k(k+3)^(2)+4)/(4(k+1)(k+2)(k+3))=((k^(3)+6k^(2)+9k+4))/(4(k+1)(k+2)(k+3)) ` ` = ((k+1)(k+1)(k+4))/(4(k+1)(k+2)(k+3))= ((k+1)(k+4))/(4(k+2)(k=3))`. `:." " P(k+1): 1/(123)+1/(234)+...+1/((k+1)(k+2)(k+3))=((k+1)(k+4))/(4(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 `1/(123)+1/(234)+1/(345)+...+1/(n(n+1)(n+2)) = (n(n+3))/(4(n+1)(n+2)) " for all values of " n in N`.