Q. 24

# Find the coefficient of x5 in the product (1 + 2x)6 (1 – x)7 using binomial theorem.

We know that- Hence   = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6

Thus, (a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6

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 + 60x2 + 160x3 + 240x4 + 192x5 + 64x6

Similarly,     = a7 + 7a6b + 21a5b2 + 35a4b3 + 35a3b4 + 21a2b5 + 7ab6 +b7

Thus, (a + b)7 = a7 + 7a6b + 21a5b2 + 35a4b3 + 35a3b4 + 21a2b5 + 7ab6 +b7

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 + 21x2 - 35x3 + 35x4 - 21x5 + 7x6 - x7

Now,

(1+2x)6(1-x)7

=(1 + 12x + 60x2 + 160x3 + 240x4 + 192x5 + 64x6)

× (1 - 7x + 21x2 - 35x3 + 35x4 - 21x5 + 7x6 - x7)

Coefficient of x5 = [(1)×(-21) + (12)×(35) + (60)×(-35)

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

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

= 171

Thus, the coefficient of x5 in the expression (1+2x)6(1-x)7 is 171.

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