Question

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

Or

Show that the middle term in the expansion of $(1+2x{)}^{2n}$ is $\frac{\mathrm{1.3.5...}(2n-1)}{n!}{2}^n.x^n$.

Answer 1

We have, $(1+2x{)}^6(1-x{)}^7=[6_{C_0}+6_{C_1}(2x)+6_{C_2}(2x{)}^2+6_{C_3}(2x{)}^3+6_{C_4}(2x{)}^4+6_{C_5}(2x{)}^5+6_{C_6}(2x{)}^6\left]\right(7_{C_0}-7_{C_1}x+7_{C_2}{x}^2-7_{C_3}{x}^3+7_{C_4}{x}^4-7_{C_5}{x}^5+7_{C_6}{x}^6-7_{C_7}{x}^7)$

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

$\therefore$ The coeffiecient of $x^5$ in the product

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

$=-21+420-2100+3360-1680+192=171$

and the coefficient of $x^7$ in the product

$=1\times (-1)+12\times 7+60\times (-21)+160\times 35+240\times (-35)+192\times 21+64\times (-7)$

$=-1+84-1260+5600-8400+4032-448$

$=-393$

$\therefore$ Sum of the coefficient of $x^5$ and $x^7=171-393=-222$

or

We have $(1+x{)}^{2n}$, here the power of expansion 2n is even.

$\therefore$ Middle term is $T_{\frac{2n}{2}+1}=T_{n+1}$

Hence, the middle term, $T_{n+1}=2n_{C_n}{x}^n$

$=\frac{2n!}{n!n!}{x}^n=\frac{\mathrm{1.2.3.4...}(2n-3)(2n-2)(2n-1)\left(2n\right)}{n!n!}{x}^n$

$=\frac{\left[\mathrm{1.3.5...}(2n-3)(2n-1)\mathrm{2.4.6...}(2n-2)2n\right]x^n}{n!n!}$

$=\frac{\mathrm{1.3.5...}(2n-3)(2n-1)2^n\mathrm{1.2.3...}n.x^n}{n!n!}$

$=\frac{\mathrm{1.3.5...}(2n-3)2^n(2n-1)n!x^n}{n!n!}$

$=\frac{\mathrm{1.3.5...}(2n-3)(2n-1){.2}^n.x^n}{n!}$ Hence proved.