Question

If $(1+x–3x^2{)}^{10}=1+a_{1}x+a_2{x}^2+...+a_{20}{x}^{20}$, then $a_2+a_4+a_6+...+a_{20}$ is equal to

• A. $\frac{3^{10}+1}{2}$
• B. $\frac{3^9+1}{2}$
• C. $\frac{3^{10}-1}{2}$
• D. $\frac{3^9-1}{2}$

Answer 1

The correct option is C

$\frac{3^{10}-1}{2}$

Step 1. Expressing the Given data:

The given equation is $(1+x–3x^2{)}^{10}=1+a_{1}x+a_2{x}^2+...+a_{20}{x}^{20}$

We have to calculate the value of $a_2+a_4+a_6+..+a_{20}$

Step 2. Put $x=1$ in the given equation

$\begin{array}{rl}(1+1–3{)}^{10}& =1+a_{1}x+a_2{x}^2+...+a_{20}{x}^{20}\\ (-1{)}^{10}& =1+a_1+a_2+...+a_{20}\\ 1& =1+a_1+a_2+...+a_{20}\\ 0& =a_1+a_2+...+a_{20}\end{array}$

Step 3. Put $x=-1$ in the given equation

$\begin{array}{rl}(1-1–3{)}^{10}& =1-a_1+a_2+...+a_{20}\\ (-3{)}^{10}& =1-a_1+a_2+...+a_{20}\\ (-3{)}^{10}& =1-a_1+a_2+...+a_{20}\\ (-3{)}^{10}-1& =-a_1+a_2+...+a_{20}\end{array}$

Step 4: Add equations in step 1 and step 2

$(-3{)}^{10}-1=2(a_2+a_4+a_6+...+a_{20})$

Hence, the value of$a_2+a_4+a_6+...+a_{20}$ is $\frac{3^{10}-1}{2}$,

Hence option (C) is the correct option.