First slide

Binomial theorem for positive integral Index

Question

$Let {(1 + x + 2 x^{2})}^{20} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{40} x^{40} . Then a_{1} + a_{3} + a_{5} + \dots . + a_{37} is equal to$

Easy

A

$2^{19} (2^{20} + 21)$
B

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

$2^{20} (2^{20} + 21)$
D

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

Solution

${(1 + x + 2 x^{2})}^{20} = a_{0} + a_{1} x + ..... a_{40} x^{40}$
$x = 1 \Rightarrow 4^{20} = a_{0} + a_{1} + \dots \dots . . + a_{40}$
$x = - 1 \Rightarrow 2^{20} = a_{0} - a_{1} + a_{2} \dots \dots . + a_{40}$
$\frac{4^{20} - 2^{20}}{20} = a_{1} + a_{3} + \dots \dots a_{37} + a_{39}$
$But a_{39} = coeff of x^{39} i n {(1 + x + 2 x^{2})}^{20}$
$= \frac{20!}{p! q! r!} 1^{p} . x^{q} {.2}^{r} . x^{2 r}$
$q + 2 r = 39, p + q + r = 20$
$q = 1, r = 19, p = 0$
$a_{39} = \frac{20!}{0! 1! 19!} {.2}^{19} = {20.2}^{19}$
$Required = \frac{2^{40} - 2^{20}}{2} - 20 . 2^{19}$
$= \frac{2^{40} - 2^{20} - {20.2}^{20}}{2}$
$= 2^{39} - 2^{19} - 20 . 2^{19} = 2^{19} (2^{20} - 21)$

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