Q3.    Find the coefficient of $x^5$ in the product $(1 + 2x)^6 (1 - x)^7$ using binomial theorem.

P Pankaj Sanodiya

First, lets expand both expressions individually,

So,

$(1+2x)^6=^6C_0+^6C_1(2x)+^6C_2(2x)^2+^6C_3(2x)^3+^6C_4(2x)^4+^6C_5(2x)^5+$$^6C_6(2x)^6$

$(1+2x)^6=^6C_0+2\times^6C_1x+4\times^6C_2x^2+8\times^6C_3x^3+16\times^6C_4x^4+32\times^6C_5x^5+$$64\times^6C_6x^6$

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

And

$(1-x)^7=^7C_0-^7C_1x+^7C_2x^2-^7C_3x^3+^7C_4x^4-^7C_5x^5+^7C_6x^6-^7C_7x^7$

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

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)$

Now, for the coefficient of $x^5$, we multiply and add those terms whose product gives $x^5$.So,

The term which contain $x^5$are,

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

$\Rightarrow 171x^5$

Hence the coefficient of $x^5$ is 171.

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