If $2^{1998} - 2^{1997} - 2^{1996} + 2^{1995} = {K .2}^{1995}$ , then the value of $K$ is

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Solution

The correct option is A 3
$2^{1998} - 2^{1997} - 2^{1996} + 2^{1995} = k . 2^{1995}$
$2^{1995} (2^{3} - 2^{2} - 2^{1} + 1) = k . 2^{1995}$
$(2^{3} - 2^{2} - 2^{1} + 1) = k$
$(8 - 4 - 2 + 1) = k$
$k = 3$

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Question 9

If $2^{1998} - 2^{1997} - 2^{1996} + 2^{1995} = k {.2}^{1995},$ then the value of $k$ is
(a) 1
(b) 2
(c) 3
(d) 4

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

\frac{j}{k} = 32

k = \frac{3}{2}

\frac{1}{2} j

4 \sum k = 0 (\frac{3^{4 - k}}{(4 - k)!}) (\frac{x^{k}}{k!}) = \frac{32}{3},

x

k

(\frac{2}{3})^{4} \times (\frac{2}{3})^{2} = (\frac{4}{9})^{(2 k - 3)}

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