Question

lf $(1+x{)}^{2n}=\sum _{r=0}^{2n}{C}_r{x}^r$ then ${\int }_0^2(C_0+C_2+C_4+\dots \dots C_{2n})dx$ equals

• A. $2^{2n-1}$
• B. $2^{3n}$
• C. $4^n$
• D. $4^{n-1}$

Answer 1

The correct option is C $4^n$

$(1+x{)}^{2n}={\sum }_{r=0}^{2n}{C}_r{X}^r$
$(1+x{)}^{2n}=C_0+C_{1}X+C_2{X}^2+\cdots Cn{c}^n$
$2^{2n}=C_0+C_1+C_2+\cdots C_{2n}$
$0=C_0+-C_1+C_2-C_3-\cdots +C_{2n}$
$x{2}^{2x-1}=C_0+C_2+\cdots C_{2n}$
${\int }_1^2(C_0+C_2+\cdots C_{2n})dx={\int }_1^2(2^{2n}-1)dx$
$=2^{2n-1}f\left(2\right)$
$=2^{2n}$
$4^n$