A) \[\frac{(1-{{c}^{n}})(1-{{a}^{n}})}{1-ac}\]
B) \[\frac{(1+a)(1-{{c}^{n}}{{a}^{n}})}{1-ac}\]
C) \[\frac{(1+{{c}^{n}})(1+{{a}^{n}})}{1-ac}\]
D) \[\frac{(1+a)(1+{{c}^{n}}{{a}^{n}})}{1+ac}\]
Correct Answer: B
Solution :
[b] Clearly the required series is \[1+a+ca+a(ca)+c(aca)+a(caca)+........\]....to 2n terms \[=1+a+ca+c{{a}^{2}}+{{c}^{2}}{{a}^{2}}+{{c}^{2}}a{{+}^{3}}\].....to 2n terms =(\[1+ca+{{c}^{2}}{{a}^{2}}+\]?.... to n terms) + \[(a+ca+{{c}^{2}}{{a}^{3}}+.......to\,\,n\,\,terms)\] \[=\frac{1\{1-{{(ca)}^{n}}\}}{1-ca}+\frac{a\{1-{{(ca)}^{n}}\}}{1-ca}\] \[=\frac{(1+a)(1-{{c}^{n}}{{a}^{n}})}{1-ca}\]
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