Thinking of expanding?

Algebra Level 4

$\large{\displaystyle \prod_{r=0}^{15} (1-2^{r}x)}$

If the coefficient of $x^{15}$ in the product is $k$ , find $k$ .

2^{104}-2^{112}

None

2^{90}-2^{105}

2^{105}-2^{121}

2^{120}-2^{104}

1 solution

Ayush Verma
Apr 10, 2015

$let\quad \prod _{ r=0 }^{ 15 }{ \left( 1-{ 2 }^{ r }x \right) } =\sum _{ i=0 }^{ 16 }{ { a }_{ i }{ x }^{ i } } \\ \\ k={ a }_{ 15 }=\sum _{ r=0 }^{ 15 }{ \cfrac { \left( -{ 2 }^{ 0 } \right) \left( -{ 2 }^{ 1 } \right) ...\left( -{ 2 }^{ 15 } \right) }{ -{ 2 }^{ r } } } =\sum _{ r=0 }^{ 15 }{ \cfrac { { \left( -1 \right) }^{ 16 }{ 2 }^{ 0+1+...+15 } }{ -{ 2 }^{ r } } } \\ =\sum _{ r=0 }^{ 15 }{ \cfrac { -{ 2 }^{ 120 } }{ { 2 }^{ r } } = } \left( -{ 2 }^{ 120 } \right) \cfrac { 1-\cfrac { 1 }{ { 2 }^{ 16 } } }{ 1-\cfrac { 1 }{ 2 } } =\left( -{ 2 }^{ 120 } \right) \left( 2-\cfrac { 1 }{ { 2 }^{ 15 } } \right) ={ 2 }^{ 105 }-{ 2 }^{ 121 }$

Thinking of expanding?

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