Question

If coefficients of $x^{20}$ in ${(1-x+x^2)}^{20}$ and in ${(1-x-x^2)}^{20}$ are respectively $a$ and $b$, then

• A. $1=b$
• B. $a>b$
• C. $a$
• D. $a+b=0$

Answer 1

The correct option is B $a>b$

Let ${(1-x+x^2)}^{20}=a_0+a_{1}x+.....+a_{40}{x}^{40}$
$\Rightarrow {(1+x+x^2)}^{20}=a_0-a_{1}x+a_2{x}^2-......+a_{40}{x}^{40}$
Thus, coefficient of $x^{20}$ in ${(1-x+x^2)}^{20}$
$=$ coefficient of $x^{20}$ in ${(1-x+x^2)}^{20}$
Therefore, we consider the expansion
${(1-x+{px}^2)}^{20}=b_0+b_{1}x+b_2{x}^2+........+b_{40}{x}^{40}$
Dividing both the sides by $x^{20}$, we get
$b^{20}=$ coefficient of constant term in
${(\frac{1}{x}+1+px)}^{20}$
But ${(\frac{1}{x}+1+px)}^{20}=1+{{}^{20}C}_1(\frac{1}{x}+px)+{{}^{20}C}_2{(\frac{1}{x}+px)}^2+....+{{}^{20}C}_{20}{(\frac{1}{x}+px)}^{20}$
$\therefore b_{20}=1+\left({{}^{20}C}_2\right)\left({{}^{2}C}_1\right)p+\left({{}^{20}C}_4\right)\left({{}^{4}C}_2\right)p^2+......+\left({{}^{20}C}_{20}\right)\left({{}^{20}C}_{10}\right)p^{10}$
Thus,
$a={\left[b_{20}\right]}_{p=1}$ and $a={\left[b_{20}\right]}_{p=-1}$
Hence, $a>b$