Question

Let ${(1-x+x^4)}^{10}=a_0+a_{1}x+a_2{x}^2+\cdots +a_{40}{x}^{40}.$ Then the correct option(s) is (are)

• A. $a_0+a_2+a_4+.....+a_{40}=\frac{3^{10}+1}{2}$
• B. $a_0=a_{40}=1$
• C. $a_1=a_{39}$
• D. $a_{39}=0$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $a_0+a_2+a_4+.....+a_{40}=\frac{3^{10}+1}{2}$
B $a_0=a_{40}=1$
D $a_{39}=0$

${(1-x+x^4)}^{10}=a_0+a_{1}x+a_2{x}^2+\cdots +a_{40}{x}^{40}\cdots \left(1\right)$
Put $x=1,$ we get
$a_0+a_1+a_2+\cdots +a_{40}=1$
Put $x=-1,$ we get
$a_0-a_1+a_2-\cdots +a_{40}=3^{10}$
Adding both,
$a_0+a_2+a_4+\cdots +a_{40}=\frac{3^{10}+1}{2}$

Put $x=0$ in $\left(1\right),$ we get $a_0=1$

Differentiating $\left(1\right)$ w.r.t. $x$ and then putting $x=0,$
$a_1=-10$

Replace $x$ by $\frac{1}{x}$ in $\left(1\right),$
${(1-\frac{1}{x}+\frac{1}{x^4})}^{10}=a_0+\frac{a_1}{x}+\frac{a_2}{x^2}+\cdots +\frac{a_{40}}{x^{40}}\Rightarrow {(x^4-x^3+1)}^4=a_0{x}^{40}+a_1{x}^{39}+a_2{x}^{38}+\cdots +a_{40}\cdots \left(2\right)$
Putting $x=0$ in $\left(2\right),$ we get $a_{40}=1$

Differentiating $\left(2\right)$ w.r.t. $x$ and then putting $x=0,$
$a_{39}=0$