Question

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

Answer 1

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

Now, it can be expanded by binomial exponension

as:

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

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

Now on multiplying $x^5$ can be finding in many ways, writing all of then below:-

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

${=}^{-7}{C}_5{x}^2+12x^7{C}_4{x}^5-30x^7{C}_3{x}^5+160{\times }^7{C}_2{x}^5-240{\times }^7{C}_1{x}^5+192x^5$

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

$=121$

$\therefore$ coefficient of $x^5=1221$