A) 1
B) 0
C) \[\infty \]
D) 4
Correct Answer: D
Solution :
Let, \[S=1+\frac{1.3}{6}+\frac{1.3.5}{6.8}+...\infty \] Þ \[\frac{S}{4}=\frac{1}{4}+\frac{1.3}{4.6}+\frac{1.3.5}{4.6.8}+....\infty \] Þ \[\frac{1}{2}-\frac{S}{8}=\frac{1}{2}-\frac{1}{2}.\frac{1}{4}-\frac{1}{2}.\frac{1.3}{4.6}-\frac{1}{2}\frac{1.3.5}{4.6.8}-....\infty \] Þ \[\frac{1}{2}-\frac{S}{8}=1-\frac{1}{2}+\frac{\frac{1}{2}\left( \frac{1}{2}-1 \right)}{1.2}-\]\[\frac{\frac{1}{2}\left( \frac{1}{2}-1 \right)\,\left( \frac{1}{2}-2 \right)}{1.2.3}\]\[+\frac{\frac{1}{2}\left( \frac{1}{2}-1 \right)\,\,\left( \frac{1}{2}-2 \right)\,\,\left( \frac{1}{2}-3 \right)}{1.2.3.4}....\infty \] Þ \[1/2-S/8={{(1-1)}^{1/2}}=0\] Þ \[S/8=1/2\Rightarrow S=4\].You need to login to perform this action.
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