Question
If 2n-1 + 2n+1 = 320, then the value of n is = ?
Answer: Option D
$$\eqalign{
& {\text{ }}{{\text{2}}^{n - 1}}{\text{ + }}{{\text{2}}^{n + 1}}{\text{ = 320}} \cr
& \Rightarrow {\text{ }}{{\text{2}}^{n - 1}}\left( {1 + {2^2}} \right){\text{ = 320}} \cr
& \Rightarrow {\text{ }}{{\text{2}}^{n - 1}} \times {\text{5 = 320}} \cr
& \Rightarrow {\text{ }}{{\text{2}}^{n - 1}}{\text{ = }}\frac{{320}}{5}{\text{ = 64}} \cr
& \Rightarrow {\left( 2 \right)^{n - 1}} = {\left( 2 \right)^6} \cr
& \Rightarrow n - 1 = 6 \cr
& \Rightarrow n = 7 \cr} $$
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$$\eqalign{
& {\text{ }}{{\text{2}}^{n - 1}}{\text{ + }}{{\text{2}}^{n + 1}}{\text{ = 320}} \cr
& \Rightarrow {\text{ }}{{\text{2}}^{n - 1}}\left( {1 + {2^2}} \right){\text{ = 320}} \cr
& \Rightarrow {\text{ }}{{\text{2}}^{n - 1}} \times {\text{5 = 320}} \cr
& \Rightarrow {\text{ }}{{\text{2}}^{n - 1}}{\text{ = }}\frac{{320}}{5}{\text{ = 64}} \cr
& \Rightarrow {\left( 2 \right)^{n - 1}} = {\left( 2 \right)^6} \cr
& \Rightarrow n - 1 = 6 \cr
& \Rightarrow n = 7 \cr} $$
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