A) \[\frac{1}{1+a}\]
B) \[\frac{2}{1+a}\]
C) \[\infty \]
D) None of these
Correct Answer: A
Solution :
\[\frac{1}{(1+a)(2+a)}\] \[+\frac{1}{(2+a)(3+a)}+\frac{1}{(3+a)(4+a)}+.....+\infty \] n th term of series \[{{T}_{n}}\]\[=\frac{1}{(n+a)(n+1+a)}\]\[=\frac{1}{n+a}-\frac{1}{n+1+a}\] \[{{T}_{1}}=\frac{1}{1+a}-\frac{1}{2+a}\];\[{{T}_{2}}=\frac{1}{2+a}-\frac{1}{3+a}\], \[{{T}_{3}}=\frac{1}{3+a}-\frac{1}{4+a}\] ..................... ..................... \[{{T}_{n-1}}=\frac{1}{n-1+a}-\frac{1}{n+a}\], \[{{T}_{n}}=\frac{1}{n+a}-\frac{1}{n+1+a}\] \\[{{S}_{n}}={{T}_{1}}+{{T}_{2}}+{{T}_{3}}+......+{{T}_{n}}\] \[=\frac{1}{1+a}-\frac{1}{n+1+a}\]\[=\frac{n}{(1+a)(n+1+a)}\] \[{{S}_{n}}=\frac{1}{(1+a)\left( 1+\frac{1}{n}+\frac{a}{n} \right)}\] \[{{S}_{\infty }}={{S}_{n}}\], when \[n\to \infty \] \ \[{{S}_{\infty }}\]\[=\frac{1}{(1+a)}\].
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