By induction for all n ∈ N \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{n(n+1)(n+2)}\) is equal to:-

Concept:

• Mathematical induction: It is a technique of proving a statement, theorem, or formula which is assumed to be true, for every natural number n.
• By generalizing this in the form of a principle that we would use to prove any mathematical statement is called the principle of mathematical induction.

Calculations:

Consider

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{n(n+1)(n+2)}\)

Clearly, the r^th term from the above series is 

Let the r^th term be \(u_r=\frac{1}{r(r+1)(r+2)}\)   ....(1)

now multiply and divide by 2 in (1)

⇒ \(u_r=\frac{1\times 2}{2r(r+1)(r+2)}\)

⇒ \(u_r=\frac{2}{2r(r+1)(r+2)}\)

add and subtract r in the numerator

⇒ \(u_r=\frac{(r+2) - r}{2r(r+1)(r+2)}\)

⇒ \(u_r=\frac{1}{2}\big[\frac{(r+2) }{r(r+1)(r+2)}- \frac{r }{r(r+1)(r+2)}\big]\)

⇒ \(u_r=\frac{1}{2}\big[\frac{1}{r(r+1)}- \frac{1}{(r+1)(r+2)}\big]\)   (2)

Now put r = 1 in (2), then we have

\(u_1=\frac{1}{2}\big[\frac{1}{1.2}- \frac{1}{2.3}\big]\)     

put r = 2 in (2)

⇒ \(u_2=\frac{1}{2}\big[\frac{1}{2.3}- \frac{1}{3.4}\big]\)     

put r = 3 in (2)

⇒ \(u_3=\frac{1}{2}\big[\frac{1}{3.4}- \frac{1}{4.5}\big]\)

      .  .  .  .  .  .  . 

put r = n in (2) 

\(u_r=\frac{1}{2}\big[\frac{1}{n(n+1)}- \frac{1}{(n+1)(n+2)}\big]\)        

Now, add all these equations and we get

\(S_n=\displaystyle\sum_{r=1}^{n} u_r=\frac{1}{2}\bigg[\frac{1}{1.2}-\frac{1}{(n+1)(n+2)}\bigg ]\)

\(=\frac{1}{4(n+1)(n+2)}\big[(n+1)(n+2)-2\big ]\)

\(=\frac{1}{4(n+1)(n+2)}\big[(n^2+n+2n+2-2\big ]\)

\(=\frac{1}{4(n+1)(n+2)}\big[n^2+3n\big ]\)

\(=\frac{n(n+3)}{4(n+1)(n+2)}\)

Hence, the correct answer is option 4).