Question

If 2^2008 – 2^2007 - 2^2006 + 2^2005 = k * 2^2005 then the value of k is equal to​

Answer 1

Answer:

k = 3

Step-by-step explanation:

$2^{2008}-2^{2007}-2^{2006}+2^{2005}=k \times 2^{2005}$

$=> (2^{2005} \times 2^3) - (2^{2005} \times 2^2) - (2^{2005} \times 2^1) + 2^{2005} = k \times 2^{2005}$

• Take $2^{2005}$ common from all the terms in L.H.S.

$=> 2^{2005}(2^3 - 2^2 - 2^1 + 1) = k \times 2^{2005}$

• Divide both the sides by $2^{2005}$.

$=> \frac{2^{2005}(2^3 - 2^2 - 2^1 + 1)}{2^{2005}} = \frac{k \times 2^{2005}}{2^{2005}}$

$=> 2^3 - 2^2 - 2^1 + 1 = k$

• Expand the terms in L.H.S.

$=> 8 - 4 - 2 + 1 = k$

$=> k = 3$

Answer 2

Step-by-step explanation:

Maths

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if 2

2008

−2

2007

−2

2006

+2

2005

=k ⋅2

2005

then the value of k is equal to

Answer

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

⇒3(2

2006

−2

2005

)=k⋅2

2005

⇒3[2

2005

(2−1)]=k⋅2

2005

⇒3(2

2005

)=k⋅2

2005

On comparing, we get

k=3

Hence, Option B is correct.