Q. 24

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

Answer :

We know that-

Hence

= a^{6} + 6a^{5}b + 15a^{4}b^{2} + 20a^{3}b^{3} + 15a^{2}b^{4} + 6ab^{5} + b^{6}

Thus, (a + b)^{6} = a^{6} + 6a^{5}b + 15a^{4}b^{2} + 20a^{3}b^{3} + 15a^{2}b^{4} + 6ab^{5} + b^{6}

Putting a = 1 & b = 2x, we get-

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

+ 15(1)^{2}(2x)^{4} + 6(1)(2x)^{5} + (2x)^{6}

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

Similarly,

= a^{7} + 7a^{6}b + 21a^{5}b^{2} + 35a^{4}b^{3} + 35a^{3}b^{4} + 21a^{2}b^{5} + 7ab^{6} +b^{7}

Thus, (a + b)^{7} = a^{7} + 7a^{6}b + 21a^{5}b^{2} + 35a^{4}b^{3} + 35a^{3}b^{4} + 21a^{2}b^{5} + 7ab^{6} +b^{7}

Putting a = 1 & b = -x, we get-

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

+ 35(1)^{3}(-x)^{4} + 21(1)^{2}(-x)^{5} + 7(1)(-x)^{6} + (-x)7

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

Now,

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

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

× (1 - 7x + 21x^{2} - 35x^{3} + 35x^{4} - 21x^{5} + 7x^{6} - x^{7})

Coefficient of x^{5} = [(1)×(-21) + (12)×(35) + (60)×(-35)

+ (160)×(21) + (240)×(-7) + (192)×(1)]

= [-21 + 420 - 2100 + 3360 - 1680 + 192]

= 171

Thus, the coefficient of x^{5} in the expression (1+2x)^{6}(1-x)7 is 171.

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