Question

If $(1+x+2x^2{)}^{20}=a_0+a_1+a_2{x}^2+...+a_{40}{x}^{40},then{a}_1+a_3+a_5+...+a_{37}equals$ :

• A. 2¹⁹(2²⁰-21)
• B. 2²⁰(2¹⁹-19)
• C. 2¹⁹(2²⁰+21)
• D. None of these

Answer 1

The correct option is A

2¹⁹(2²⁰-21)

$(1+x+2x^2{)}^{20}=a_0+a_{1}x+a_2{x}^2+....+a_{40}{x}^{40}$

x = 1 , x = -1 gives

$a_0+a_1+a_2+...+a_{40}=4^{20}$

$a_0+a_1+a_2+...+a_{40}=4^{20}$

$\Rightarrow$ $a_1+a_3+a_5+....+a_{39}=(4^{20}-2^{20})$

$\Rightarrow$ $a_1+a_3+....+a_{37}=2^{19}(2^{20}-1)-a_{39}.$

Now, $a_{39}$ = coeff. of $x^{39}$ in [1 + x(1 + 2x^2 )]^{20} \)

= coeff. of $x^{39}$ in [1 + x(1 + 2x )]^{20} \)

= coeff. of $x^{39}$ in ${}^{2}0C_{20}.x^{20}(1+2x{)}^{20}.$

= coeff. of $x^{19}in(1+2x{)}^{20}$

= ${}^{20}{C}_{19}{2}^{19}=20\times 2^{19}$

Hence , $a_1+a_3+....+a_{37}=2^{19}(2^{20}-1)-20\times 2^{19}$

= $2^{19}(2^{20}-21)$ .