Let (1+x+2x^2)^{20} = a_0 + a_1x + a_2x^2 + l | vedclass

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(B) Given $(1+x+2x^2)^{20} = \sum_{k=0}^{40} a_k x^k$.Let $f(x) = (1+x+2x^2)^{20}$.$f(1) = a_0 + a_1 + a_2 + \ldots + a_{40} = (1+1+2)^{20} = 4^{20} =

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

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(B) Given $(1+x+2x^2)^{20} = \sum_{k=0}^{40} a_k x^k$.Let $f(x) = (1+x+2x^2)^{20}$.$f(1) = a_0 + a_1 + a_2 + \ldots + a_{40} = (1+1+2)^{20} = 4^{20} = 2^{40}$.$f(-1) = a_0 - a_1 + a_2 - \ldots + a_{40} = (1-1+2)^{20} = 2^{20}$.Subtracting the two equations: $f(1) - f(-1) = 2(a_1 + a_3 + \ldots + a_{39}) = 2^{40} - 2^{20}$.So,$a_1 + a_3 + \ldots + a_{39} = \frac{2^{40} - 2^{20}}{2} = 2^{39} - 2^{19}$.We need $a_1 + a_3 + \ldots + a_{37} = (a_1 + a_3 + \ldots + a_{39}) - a_{39}$.To find $a_{39}$,we look at the coefficient of $x^{39}$ in $(1+x+2x^2)^{20}$.Using the multinomial expansion,the term $x^{39}$ is obtained from $\frac{20!}{n_1! n_2! n_3!} (1)^{n_1} (x)^{n_2} (2x^2)^{n_3}$ where $n_1+n_2+n_3=20$ and $n_2+2n_3=39$.If $n_3=20$,$n_2=-1$ (impossible). If $n_3=19$,$n_2=1$,then $n_1=0$.Coefficient $a_{39} = \frac{20!}{0! 1! 19!} (1)^0 (1)^1 (2)^{19} = 20 	imes 2^{19}$.Thus,$a_1 + a_3 + \ldots + a_{37} = 2^{39} - 2^{19} - 20 	imes 2^{19} = 2^{39} - 21 	imes 2^{19} = 2^{19}(2^{20} - 21)$.

Let $(1+x+2x^2)^{20} = a_0 + a_1x + a_2x^2 + \ldots + a_{40}x^{40}$,then $a_1 + a_3 + a_5 + \ldots + a_{37}$ is equal to

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