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If 2^n-1 + 2ⁿ⁺¹ = 320, then the value of n is = ?

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} 2^{n - 1} + 2^{n + 1} = 320 \\ \Rightarrow 2^{n - 1} (1 + 2^{2}) = 320 \\ \Rightarrow 2^{n - 1} \times 5  = 320 \\ \Rightarrow 2^{n - 1} = \frac{320}{5} = 64 \\ \Rightarrow {(2)}^{n - 1} = {(2)}^{6} \\ \Rightarrow n - 1 = 6 \\ \Rightarrow n = 7 \end{aligned}

$\eqalign{ & {\text{ }}{{\text{2}}^{n - 1}}{\text{ + }}{{\text{2}}^{n + 1}}{\text{ = 320}} \cr & \Rightarrow {\text{ }}{{\text{2}}^{n - 1}}\left( {1 + {2^2}} \right){\text{ = 320}} \cr & \Rightarrow {\text{ }}{{\text{2}}^{n - 1}} \times {\text{5 = 320}} \cr & \Rightarrow {\text{ }}{{\text{2}}^{n - 1}}{\text{ = }}\frac{{320}}{5}{\text{ = 64}} \cr & \Rightarrow {\left( 2 \right)^{n - 1}} = {\left( 2 \right)^6} \cr & \Rightarrow n - 1 = 6 \cr & \Rightarrow n = 7 \cr}$

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", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} 2^{n - 1} + 2^{n + 1} = 320 \\ \Rightarrow 2^{n - 1} (1 + 2^{2}) = 320 \\ \Rightarrow 2^{n - 1} \times 5  = 320 \\ \Rightarrow 2^{n - 1} = \frac{320}{5} = 64 \\ \Rightarrow {(2)}^{n - 1} = {(2)}^{6} \\ \Rightarrow n - 1 = 6 \\ \Rightarrow n = 7 \end{aligned}

$\eqalign{ & {\text{ }}{{\text{2}}^{n - 1}}{\text{ + }}{{\text{2}}^{n + 1}}{\text{ = 320}} \cr & \Rightarrow {\text{ }}{{\text{2}}^{n - 1}}\left( {1 + {2^2}} \right){\text{ = 320}} \cr & \Rightarrow {\text{ }}{{\text{2}}^{n - 1}} \times {\text{5 = 320}} \cr & \Rightarrow {\text{ }}{{\text{2}}^{n - 1}}{\text{ = }}\frac{{320}}{5}{\text{ = 64}} \cr & \Rightarrow {\left( 2 \right)^{n - 1}} = {\left( 2 \right)^6} \cr & \Rightarrow n - 1 = 6 \cr & \Rightarrow n = 7 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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