First slide

Binomial theorem for positive integral Index

Question

${(1 + 2 x - 3 x^{2})}^{10} = a_{0} + a_{1} x + a_{2} x^{2} + \dots . a_{20} x^{20} then the value of$ $a_{0} + a_{1} + a_{2} + \dots a_{20} is$

Very Easy

A

$6^{20}$
B

$2^{20}$
C

0
D

1

Solution

In the given equation put x = 1
${(1 + 2 - 3)}^{20} = a_{0} + a_{1} + a_{2} + .... + a_{20}$
$\Rightarrow a_{0} + a_{1} + a_{2} + .... + a_{20} = 0$

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