Question

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

Answer 1

By using the formula :

$(a+b{)}^n=n_{C_0}{a}^n+n_{C_1}{a}^{n-1}b+n_{C_2}{a}^{n-2}{b}^2+.........n_{C_r}{a}^{n-r}{b}^r+......n_{C_n}{b}^n$

$(1+2x{)}^6=6_{C_0}.1+6_{C_1}\left(2x\right)+6_{C_2}(2x{)}^2+6_{C_2}(2x{)}^3+6_{C_4}(2x{)}^4+6_{C_6}(2x{)}^6$

$=1+12x+60x^2+20\times 8x^3+15\times 16x^4+6\times 32x^6+64x^5$

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

$(1-x{)}^7=1-7_{C_1}x+7_{C_2}{x}^2-7_{C_3}{x}^3+7_{C_4}{x}^4-7_{C_5}{x}^5+...{.7}_{C_7}{x}^7$

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

multiply equation (i) and (ii) .

$=1\times (-21)+12\times 35+60\times (-35)+160\times 21+240\times (-7)+192\times 1$

$=-21+420-2100+3360-1680+192$

$=171$