A) \[{{8}^{n}}+1\]
B) \[{{4}^{n}}-3n-1\]
C) \[{{3}^{2n}}+3n+1\]
D) \[{{10}^{n}}+1\]
Correct Answer: B
Solution :
Now, \[{{4}^{n}}={{(1+3)}^{n}}\] \[=1+3n+\frac{n(n-1)}{2!}{{3}^{2}}+...\] \[\Rightarrow \] \[{{4}^{n}}-3n-1={{3}^{2}}\left[ \frac{n(n-1)}{2!}+... \right]\] It is dear from above that \[{{4}^{n}}-3n-1\] is divisible by 9.
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