Find the coefficient of $x^{5}$ in the product ${(1 + 2 x)}^{6} {(1 - x)}^{7}$ using binomial theorem.

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Solution

${(1 + 2 x)}^{6} = {6_{C}}_{0} + {6_{C}}_{1} 2 x + {6_{C}}_{2} {(2 x)}^{2} + {6_{C}}_{3} {(2 x)}^{3} + {6_{C}}_{4} {(2 x)}^{4} + {6_{C}}_{5} {(2 x)}^{5} + {6_{C}}_{6} {(2 x)}^{6}$

$= 1 + 6 \times 2 x + 15 \times {(2 x)}^{2} + 20 \times {(2 x)}^{3} + 15 \times {(2 x)}^{4} + 6 \times {(2 x)}^{5} + {(2 x)}^{6}$

$= 1 + 12 x + 60 x^{2} + 160 x^{3} + 240 x^{4} + 192 x^{5} + 64 x^{6}$

${(1 - x)}^{7} = {7_{C}}_{0} - {7_{C}}_{1} x + {7_{C}}_{2} x^{2} - {7_{C}}_{3} x^{3} + {7_{C}}_{4} x^{4} - {7_{C}}_{5} x^{5} + {7_{C}}_{6} x^{6} - {7_{C}}_{7} x^{7}$

$= 1 - 7 x + 21 x^{2} - 35 x^{3} + 35 x^{4} - 21 x^{5} + 7 x^{6} - x^{7}$

$∴ {(1 + 2 x)}^{6} {(1 - x)}^{7} = (1 + 12 x + 60 x^{2} + 160 x^{3} + 240 x^{4} + 192 x^{5} + 64 x^{6}) (1 - 7 x + 21 x^{2} - 35 x^{3} + 35 x^{4} - 21 x^{5} + 7 x^{6} - x^{7})$

The complete multiplication of the two bracets is not required to be carried out.Only those terms, which involve $x^{5}$ are required.
The terms containing $x^{5}$ are

$= 1 (- 21 x^{5}) + (12 x) (35 x^{4}) + (60 x^{2}) (- 35 x^{3}) + (160 x^{3}) (21 x^{2}) + (240 x^{4}) (- 7 x) + (192 x^{5}) (1)$

$= 171 x^{5}$ on simplification

Thus, the coefficient of $x^{5}$ in the given product is $171$

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Similar questions

Find the coefficient of $x^{5}$ in ${(1 + 2 x)}^{6} + {(1 - x)}^{7}$ .

x^{8}

{(1 + 2 x)}^{6} {(1 - x)}^{7}

a^{4}

(1 + 2 a)^{4} (2 - a)^{5}

a^{4}

(1 + 2 a)^{4} (2 - a)^{5}

x^{5}

(1 + 2 x)^{6} (1 - x)^{7}

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