First slide

Binomial theorem for positive integral Index

Question

The sum of the coefficient of all odd degree terms in the expansion of ${(x + \sqrt{x^{3} - 1})}^{5} + {(x - \sqrt{x^{3} - 1})}^{5}, (x > 1)$

Easy

A

0
B

1
C

2
D

-1

Solution

$We know that (x + a)^{5} + (x - a)^{5} = 2 [{}^{5}C_{0} x^{5} + {}^{5}C_{2} x^{3} a^{2} + {}^{5}C_{4} x a^{4}]$
$∴$ ${(x + \sqrt{x^{3} - 1})}^{5} + {(x - \sqrt{x^{3} - 1})}^{5}$
$\begin{array}{l} = 2 [{}^{5}C_{0} x^{5} + {}^{5}C_{2} x^{3} (x^{3} - 1) + {}^{5}C_{4} x {(x^{3} - 1)}^{2}] \\ = 2 [{}^{5}C_{0} x^{5} + {}^{5}C_{2} (x^{6} - x^{3}) + {}^{5}C_{4} x (x^{6} - 2 x^{3} + 1)] \end{array}$
Thus, the sum of coefficient of all odd degree terms
$\begin{array}{l} = 2 [{}^{5}C_{0} - {}^{5}C_{2} + {}^{5}C_{4} + {}^{5}C_{4}] \\ = 2 [1 - 10 + 5 + 5] = 2 \end{array}$

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