Using the principle of mathematical 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`

Explanation

We have,
`P(n) = (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 `n = 1`,
L.H.S ` = (1)/(1.2.3) = 1/6` and R.H.S. `= (1.4)/(4.2.3) = (1)/(6)`
Thus, `P(1)` is true.
Let `P(n)` be true for some ` n = k`.
Then `(1)/(1.2.3) + (1)/(2.3.4) + (1)/(3.4.5) + "......." + (1)/(k(k+1)(k+2))`
`= (k(k+3))/(4(k+1)(k+2))"....."(1)`
Adding `(1)/((k+1)(k+3))` on both sides, we get
`(1)/(2.3.4) + (1)/(2.3.4) + (1)/(3.4.5) + "......" + (1)/(k(k+1)(k+2)) + (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))`
`= (k(k+3))/((k+1)(k+2)) + (1)/((k+1)(k+2)(k+3))`
`= (1)/((k+1)(k+2)) {(k(k+3))/(4) + (1)/(k+3)}`
`= (1)/((k+1)(k+2)) {(k(k+3)^(2) + 4)/(4(k+3))}`
`= (1)/((k+1)(k+2)){(k^(3)+6k^(2) + 9k +4)/(4(k+3))}`
`= ((k+1)^(2)(k+4))/(4(k+1)(k+2)(k+3))`
`= ((k +1){(k+1)+3})/(4{(k+1)+1}{(k+1)+2)})`
Thus, `P(k+1)` is true whenever `P(k)` is true.
Hence, by the principle of mathematical induction, statement
`P(n)` is true for all natural numbers.