Lf 1-x-x220-r-040arxr- then a2-2a4-3a6-20a40-

The correct option is D 10The value of a_2+2a_4+3a_6...+20a_40 It is given by, f'(1) f'( 1)4 ...(i)Since (1 x x^2)^20=∑ _r=0 ^40 a_rx^r f(x ...

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Variable Separable Method

lf 1-x-x220...

Question

lf $(1 - x - x^{2})^{20} = 40 \sum r = 0 a_{r} x^{r}$ , then $a_{2} + 2 a_{4} + 3 a_{6} + \dots \dots \dots \dots + 20 a_{40} =$

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Similar questions

x^{20}

{(1 - x + x^{2})}^{20}

{(1 - x - x^{2})}^{20}

a

b

(1 + x + x^{2})^{20} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{40} x^{40},

a_{0} + a_{1} + a_{2} + \dots + a_{19}

Let x \in R

P = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 0 & 2 & 2 0 & 0 & 3 \end{matrix} ⎤ ⎥ ⎦,

Q = ⎡ ⎢ ⎣ \begin{matrix} 2 & x & x 0 & 4 & 0 x & x & 6 \end{matrix} ⎤ ⎥ ⎦

R = P Q P^{- 1} .

{(1 + x + x^{2})}^{20} = 40 \sum r = 0 a_{r} x^{r}

a_{1} + a_{3} + a_{5} + \dots a_{37}

2 {sinh}^{- 1} (\frac{a}{\sqrt{1 - a^{2}}}) = log (\frac{1 + x}{1 - x})

x =

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