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

Question:

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

Solution:

Using Binomial Theorem, the expressions, $(1+2 x)^{6}$ and $(1-x)^{7}$, can be expanded as

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

$+{ }^{6} \mathrm{C}_{5}(2 x)^{5}+{ }^{6} \mathrm{C}_{6}(2 x)^{6}$

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

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

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

$-{ }^{7} \mathrm{C}_{5}(x)^{5}+{ }^{7} \mathrm{C}_{6}(x)^{6}-{ }^{7} \mathrm{C}_{7}(x)^{7}$

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

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

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

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

The terms containing $x^{5}$ are

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

$=171 x^{3}$

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