Question

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

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

Answer 1

Answer: (D)

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