What is expand (2^i(2^{k-i}-1)^2)/(2^k-1) ?

The solution to expand (2^i(2^{k-i}-1)^2)/(2^k-1) is (2^{2k-i}-2^{k+1}+2^i)/(2^k-1)

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expand (2^i(2^{k-i}-1)^2)/(2^k-1)

expand

\frac{2 ^{i} ( 2 ^{k - i} - 1 ) ^{2}}{2 ^{k} - 1}

\frac{2 ^{2 k - i} - 2 ^{k + 1} + 2 ^{i}}{2 ^{k} - 1}

Solution steps

\frac{2 ^{i} ( 2 ^{k - i} - 1 ) ^{2}}{2 ^{k} - 1}

(2^{k - i} - 1)^{2} = 4^{k - i} - 2^{1 + k - i} + 1

= \frac{2 ^{i} ( 4 ^{k - i} + 1 - 2 ^{k + 1 - i} )}{2 ^{k} - 1}

Expand

2^{i} (4^{k - i} - 2^{1 + k - i} + 1) : 2^{2 k - i} - 2^{k + 1} + 2^{i}

= \frac{2 ^{2 k - i} - 2 ^{k + 1} + 2 ^{i}}{2 ^{k} - 1}

12 (\frac{2}{3} x + 7)

(a + b c) (a - b c)

(4 - x 7)^{2}

\frac{3}{( x + 2 ) ^{2}}

(cos (t) + sin (t)) \cdot (- sin (t))

What is expand (2^i(2^{k-i}-1)^2)/(2^k-1) ?
The solution to expand (2^i(2^{k-i}-1)^2)/(2^k-1) is (2^{2k-i}-2^{k+1}+2^i)/(2^k-1)