Question

If $(2x^2 - x - 1)^5 = a_0 + a_1x + a_2x^2 + ... + a_{10}x^{10}$, then, $a_2 + a_4 + a_6 + a_8 + a_{10} =$

• A. $15$
• B. $30$
• C. $16$
• D. $17$

Answer 1

Answer: (D)

$17$

Answer 2

$(2x^2 - x - 1)^5 = a_0 + a_1x + a_2x^2 + ... + a_{10}x^{10}$ Put $x = 1$, we get, $0 = a_0 + a_1 + a_2 + ... + a_{10} \quad ...(i)$ Put $x = - 1$, we get, $32 = a_0 - a_1 + a_2 - ... + a_{10} \quad ...(ii)$ Adding $(i)$ and $(ii)$, we get $a_2 + a_4 + ... + a_{10} = (32/2) + 1$ $= 16 + 1 = 17 \,\,(\because a_0 = - 1)$