Question

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

Answer 1

Given equation :- $(1+2x{)}^6(1-x{)}^7$

Now, it can be expanded by binomial expansion

$(1+2x{)}^6={(}^6{C}_0{1}^6(2x{)}^0{+}^6{C}_1{1}^5\left(2x{)}^1{+}^6{C}_2\right(2x{)}^2{+}^6{C}_3\left(2x{)}^3{+}^6{C}_4\right(2x{)}^4{+}^6{C}_5(2x{)}^5....)$

$(1-x{)}^7={(}^7{C}_0{1}^7(-x{)}^0{+}^7{C}_1{1}^6(-x{)}^1{+}^7{C}_2{1}^5(-x{)}^2{+}^7{C}_3{1}^4(-x{)}^3{+}^7{C}_4{1}^3(-x{)}^4{+}^7{C}_5(-x{)}^5)$

Now on multiplying $x^5$ can be finded in many ways sorting all of then below:-

${}^6{C}_0{1}^6\left(2x{)}^0{\times }^7{C}_5\right(-x{)}^5{+}^6{C}_1\left(2x\right){\times }^7{C}_4{x}^4{+}^6{C}_2\left(2x{)}^2{\times }^7{C}_3\right(-x{)}^3{+}^6{C}_3\left(2x{)}^3{\times }^7{C}_2\right(-x{)}^2{+}^6{C}_4\left(2x{)}^4{}^{}{\times }^7{C}_1\right(-x\left){+}^6{C}_2\right(2x{)}^5$

${=}^{-7}{C}_5{x}^5+12{\times }^7{C}_4{x}^5-30{\times }^7{C}_3{x}^5+160{\times }^7{C}_2{x}^5-240{\times }^7{C}_1{x}^5+192x^5$

$=-21x^5+420x^5-1050x^5+3360x^5-1680x^5+192x^5$

$=1221$

$\therefore Coeff.of{x}^5=1221$