Question

The sum of the coefficients of all the even powers of $x$ in the expansion of $(2x^2–3x+1{)}^{11}$ is $P_1^m{P}_2^n,$ where $P_1$ and $P_2$ are prime numbers such that $P_1<P_2$ and $m$ and $n$ are positive integers. Then the value of $\frac{P_1+P_2}{n-m}$ is

• A. $2$
• B. $5$
• C. $-2$
• D. $7$

Answer 1

The correct option is B $5$

Let $(2x^2–3x+1{)}^{11}=a_0+a_{1}x+a_2{x}^2+\dots a_{22}{x}^{22}\cdots (1)$
Put $x=1$ in equation $\left(1\right),$ we get
$(2-3+1{)}^{11}=a_0+a_1+a_2+a_3+\dots +a_{22}$
$\Rightarrow a_0+a_1+a_2+a_3+\dots +a_{22}=0\cdots \left(2\right)$
Again put $x=-1$ in equation $\left(1\right),$ we get
$(2+3+1{)}^{11}=a_0-a_1+a_2-a_3+\dots a_{22}$
$\Rightarrow a_0-a_1+a_2-a_3+\dots +a_{22}=6^{11}\cdots \left(3\right)$
Adding equations $\left(2\right)$ and $\left(3\right),$ we get
$2(a_0+a_2+a_4+\dots +a_{22})=6^{11}$
$\Rightarrow a_0+a_2+a_4+\dots +a_{22}=\frac{2^{11}\cdot 3^{11}}{2}$
$\Rightarrow a_0+a_2+a_4+\dots +a_{22}=2^{10}\cdot 3^{11}=P_1^m{P}_2^n$
$\therefore P_1=2,P_2=3,m=10$ and $n=11$
Hence, the value of $\frac{P_1+P_2}{n-m}$ is $5.$