Question

Find the coefficient of a4 in the product $(1+2a{)}^4(2-a{)}^5$ using binomial theorem.

Answer 1

We have,

$(1+2a{)}^4(2-a{)}^5$

Now,

$(1+2a{)}^4={}^4{C}_0+{}^4{C}_{1}2a+{}^4{C}_2(2a{)}^2+{}^4{C}_3(2a{)}^3+{}^4{C}_4(2a{)}^4$

and, $(2-5{)}^5={}^5{C}_0\times 2^5+{}^5{C}_2\times 2^4(-a)+{}^5{C}_2\times 2^3(-a{)}^3+{}^5{C}_3\times 2^2(-a{)}^3+{}^5{C}_4\times 2(-a{)}^4+{}^5{C}_5(-a{)}^5$

$={}^5{C}_0\times 2^5-{}^5{C}_1\times 2^4\times a+{}^5{C}_2\times 2^3{a}^2-{}^5{C}_3\times 2^2\times a^3+{}^5{C}_4\times 2\times a^4-{}^5{C}_5\times a^5$

$\therefore (1+2a{)}^4(2-a{)}^5=[{}^4{C}_0+{}^4{C}_{1}2a+{}^4{C}_2(2a{)}^2+{}^4{C}_3(2a{)}^3+{}^4{C}_4(2a{)}^4\left]\right[{}^5{C}_0\times 2^5-{}^5{C}_1\times 2^4\times a+{}^5{C}_2\times 2^3\times a^2-{}^5{C}_3\times 2^2\times a^3+{}^5{C}_4\times 2\times a^4-{}^5{C}_5\times a^5]$

$\therefore$ Coefficients of $a^4=2{}^5{C}_4-{}^4{C}_1\times 2\times {}^5{C}_3\times 2^2+{}^4{C}_2(2{)}^2\times {}^5{C}_2\times 2^3-{}^4{C}_3(2{)}^3\times {}^5{C}_1\times 2^4+{}^5{C}_0\times 2^5$

$=2\times 5-8\times 4\times 10+32\times 6\times 10-128\times 4\times 512\times 1\times 1$

=10-320+1920-2560+512

=2442-2880

=-438

$\therefore$ Coefficients of $a^4=-438.$