Answer:
Option (a) [tex]\sum_{k=1}^{\infty} \dfrac{3}{k(k+1)}[/tex]
Step-by-step explanation:
The given expression is :
[tex]\dfrac{3}{1{\cdot} 2}+\dfrac{3}{2{\cdot} 3}+\dfrac{3}{3{\cdot} 4}+\dfrac{3}{4{\cdot} 5}+....[/tex]
We need to rewrite the sum using sigma notation.
The numerator is 3 in all terms.
At denominator, in first term (1)(2), in second term (2)(3) and so on.
A general term for the denominator is k(k+1).
Sigma notation is :
[tex]\sum_{k=1}^{\infty} \dfrac{3}{k(k+1)}[/tex]
Hence, the correct option is (A).
