A) 3
B) 5
C) 7
D) None of these
Correct Answer: C
Solution :
We have \[1+a+{{a}^{2}}+.......+{{a}^{x}}=(1+a)(1+{{a}^{2}})(1+{{a}^{4}})\] \[\Rightarrow \]\[\frac{(1-{{a}^{x+1}})}{(1-a)}=(1+a)(1+{{a}^{2}})+(1+{{a}^{4}})\] \[\Rightarrow \]\[(1-{{a}^{x+1}})=(1-a)(1+a)(1+{{a}^{2}})(1+{{a}^{4}})\] \[\Rightarrow \]\[(1-{{a}^{x+1}})=(1-{{a}^{8}})\]\[\Rightarrow \]\[x+1=8\]\[\Rightarrow \]\[x=7\].
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