If 22008-22007-22006-22005-k -22005 then the value of k is equal to

The correct option is B 3 2^2008 2^2007 2^2006+2^2005=k· 2^2005 ⇒ [2^2008 2^2006] [2^2007 2^2005]=k· 2^2005 ⇒ 2^2006(2^2 1) 2^2005(2^2 1)=k·2^2005 ...

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Standard XII

Mathematics

Method of Difference

if 22008-22...

Question

if $2^{2008} - 2^{2007} - 2^{2006} + 2^{2005} = k \cdot 2^{2005}$ then the value of $k$ is equal to

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Solution

The correct option is B 3
$2^{2008} - 2^{2007} - 2^{2006} + 2^{2005} = k \cdot 2^{2005}$
$\Rightarrow [2^{2008} - 2^{2006}] - [2^{2007} - 2^{2005}] = k \cdot 2^{2005}$
$\Rightarrow 2^{2006} (2^{2} - 1) - 2^{2005} (2^{2} - 1) = k \cdot 2^{2005}$
$\Rightarrow 3 (2^{2006} - 2^{2005}) = k \cdot 2^{2005}$
$\Rightarrow 3 [2^{2005} (2 - 1)] = k \cdot 2^{2005}$
$\Rightarrow 3 (2^{2005}) = k \cdot 2^{2005}$
On comparing, we get
$k = 3$
Hence, Option B is correct.

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if 22008−22007−22006+22005=k ⋅22005 then the value of k is equal to

The correct option is B 322008−22007−22006+22005=k⋅22005⇒[22008−22006]−[22007−22005]=k⋅22005⇒22006(22−1)−22005(22−1)=k⋅22005⇒3(22006−22005)=k⋅22005⇒3[22005(2−1)]=k⋅22005⇒3(22005)=k⋅22005On comparing, we getk=3Hence, Option B is correct.

if $2^{2008} - 2^{2007} - 2^{2006} + 2^{2005} = k \cdot 2^{2005}$ then the value of $k$ is equal to