Can you solve this?

The given expression can be simplified as

$2^{1995}(2^{3}-2^{2}-2+1)=k \times 2^{1995}$

$k=8-4-2+1$

$k=3$

$2^{1998}-2 ^{1997}-2^{1996}+2^{1995}=k\times 2^{1995}$ Factoring the L.H.S, We get: $2^{1995}(2^3-2^2-2+1)=k\times 2^{1995}$ Cancelling out the $2^{1995}s$ from L.H.S and R.H.S, we get: $2^3-2^2-2+1=k$ Solving,we get: $2^3-2^2-2+1=k$ $k=8-4-2+1=4-2+1=2+1=\boxed{3}$

If $2^{n+1}- 2^{n} = 2^{n}$ then:

$2^{1998} - 2^{1997} = 2^{1997}$

$2^{1997} - 2^{1996} = 2^{1996}$

If $2^{n+1} + 2^{n} = 2^{n} \times 3$ then:

$2^{1996} + 2^{1995} = 2^{1995} \times 3$

Therefore k is 3

$2^{ 1998 }-2^{ 1997 }-2^{ 1996 }+2^{ 1995 }=k\cdot 2^{ 1995 }$

We can do this:

$\left( 2^{ 1998 }-2^{ 1997 } \right) -2^{ 1996 }+2^{ 1995 }=k\cdot 2^{ 1995 }$

We know that:

$2^{ n+1 }-{ 2 }^{ n }=2\cdot { 2 }^{ n }-{ 2 }^{ n }={ 2 }^{ n }+{ 2 }^{ n }{ -2 }^{ n }={ 2 }^{ n }$

So:

$\left( 2^{ 1998 }-2^{ 1997 } \right) -2^{ 1996 }+2^{ 1995 }=k\cdot 2^{ 1995 }$

$2^{ 1997 }-{ 2 }^{ 1996 }+{ 2 }^{ 1995 }=k\cdot 2^{ 1995 }$

For the same principle:

$\left( { 2 }^{ 1997 }-{ 2 }^{ 1996 } \right) +{ 2 }^{ 1995 }={ 2 }^{ 1996 }{ +2 }^{ 1995 }$

Simplifying:

${ 2 }^{ 1996 }+{ 2 }^{ 1995 }=k\cdot { 2 }^{ 1995 }$

$2\cdot { 2 }^{ 1995 }+{ 2 }^{ 1995 }=k\cdot { 2 }^{ 1995 }$

${ 2 }^{ 1995 }+{ 2 }^{ 1995 }+{ 2 }^{ 1995 }=k\cdot { 2 }^{ 1995 }$

$3\cdot { 2 }^{ 1995 }=k\cdot { 2 }^{ 1995 }$

$\boxed{k=3}$

2^1997 - 2^1996 + 2^1995 = k 2 ^1995 2 ^1995 ( 2^3 - 2^2 - 2 + 1 ) = k 2 ^1995 3 * 2 ^1995 = k 2 ^1995, k=3

2^1997 - 2^1996 + 2^1995 = k 2 ^1995 2 ^1995 ( 2^3 - 2^2 - 2 + 1 ) = k 2 ^1995 3 * 2 ^1995 = k 2 ^1995, k=3

The given expression can be simplified

2^{1995}.2^{3}-2^{1995}.2^{2}-2^{1995}.2^{1}+2^{1995}=k.2^{1995}

2^{1995}(2^{3}-2^{2}-2^{1}+1)=k.2^{1995}

(8-4-2+1)=k

k=3

2^{1995} (2^{3}-2^{2}-2^{1}+1)=K2^{1995} 8-4-2+1=K K=3

Divide All terms by 2^1995 and take denominator to numerator to solve!!