Question 8

If $$A = 2^{32}, B = 2^{31} + 2^{30} + 2^{29} + … + 2^0$$ and $$C = 3^{15} + 3^{14} + 3^{13} + … +3^0$$, then which of the following option is TRUE?

Solution

$$B=2^{31}+2^{30}+2^{29}+\dots+2^0$$

or, $$B=2^0.\ \frac{2^{32}-1}{2-1}=\left(2^{32}-1\right)\ .$$

And , $$C = 3^{15} + 3^{14} + 3^{13} + … +3^0$$

or, $$C=3^0.\ \frac{3^{16}-1}{3-1}=\frac{1}{2}\left(3^{16}-1\ \right).$$

So, $$A>B>C\ .$$

C is correct choice.

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