Q. If $({2x}^{2} - x - {1)}^{5} = a_{0} + a_{1}x + a_{2}x^{2} + ... + a_{10}x^{10},\ then\ a_{2} + a_{4} + a_{6} + a_{8} + a_{10}$is equal to

A) 15
B) 30
C) 16
D) 32
Explanation

(a)
$({2x}^{2} - x - {1)}^{5} = a_{0} + a_{1}x + a_{2}x^{2} + \ldots + a_{10}x^{10}$
On putting $x = 0,$ we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - 1 = a_{0}$
On putting $x = 1$, we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 = a_{0} + a_{1} + a_{2} + ... + a_{10}$ …(i)
On putting $x = - 1$, we get
$(2 + 1 - 1)^{5} = a_{0} - a_{1} + a_{2} - ... + a_{10}$ …(ii)
On adding Eqs. (i) and (ii), we get
$\ \ \ \ \ \ \ 0 + (2)^{5} = 2\left( a_{0} + a_{2} + \ldots + a_{10} \right)$
$\Longrightarrow \ \ \ \ 16 - 1 = a_{2} + \ldots + a_{10}$
$\Longrightarrow a_{2} + a_{3} + ... + a_{10} = 15$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.