Let 1-x-2 x220-r-040 ar xr - then a1-a3-a5- -a39 is equal to-A- -218B- -219C- -220D- -221

The correct option is B 219 Let f(x)=(1+x 2x^2)^20 lt;/br gt; Then a_1+a_3+a_5+.......+a_39=f(1) f( 1)/2=0 2^20/2= 2^19 ...

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Let 1+x-2 x22...

Question

Let $(1 + x - 2 x^{2})^{20} = \sum_{r = 0}^{40} a_{r} x^{r} . t h e n a_{1} + a_{3} + a_{5} + . . . . + a_{39}$ is equal to:

-2¹⁹

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Similar questions

{(1 + x + 2 x^{2})}^{20} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . + a_{10} x^{40}

a_{1} + a_{3} + a_{5} + . . . . . . + a_{39}

(1 + x + 2 x^{2})^{20} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . a_{40} x^{40}

a_{1} + a_{3} + a_{5} + . . . . a_{39}

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

a_{1} + a_{3} + a_{5} + \dots + a_{39} = - 2^{k}

k =

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

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

If $(1 + x + 2 x^{2})^{20} = a_{0} + a_{1} + a_{2} x^{2} + . . . + a_{40} x^{40}, t h e n a_{1} + a_{3} + a_{5} + . . . + a_{37} e q u a l s$ :

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