Question

If $C_0,C_1,C_2,.....,C_n$ are the binomial coefficients, then $2C_1+2^3{C}_3+2^5{C}_5+...$ equals

• A. $\frac{3^n+{\left(-1\right)}^n}{2}$
• B. $\frac{3^n-{\left(-1\right)}^n}{2}$
• C. $\frac{3^n+1}{2}$
• D. $\frac{3^n-1}{2}$

Answer 1

Answer: (B)

$\frac{3^n-{\left(-1\right)}^n}{2}$

We have,

$(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+C_3{x}^3+....+C_n{x}^n...(1\left)\text{and}\right(1-x{)}^n=C_0-C_{1}x+C_2{x}^2-C_3{x}^3+.....+(-1{)}^n\cdot C_n{x}^n...(2)$

Subtracting equation $\left(2\right)$ from $\left(1\right)$

$\Rightarrow (1+x{)}^n-(1-x{)}^n=2[C_{1}x+C_3{x}^3+C_5{x}^5+....]$

$\Rightarrow \frac{1}{2}\left[\right(1+x{)}^n-(1-x{)}^n]=C_{1}x+C_3{x}^3+C_5{x}^5+.....$

Putting $x=2$, we get,

$2C_1+2^3{C}_3+2^5{C}_5+...=\frac{3^n-(-1{)}^n}{2}$

Hence, the correct answer is option (B).