Question

If $C_0,C_2,C_4,..........$ are binomial coefficients in the expansion of $(1+x{)}^9$, then $C_0+C_2+C_4+C_6+C_8=$

• A. $2^7$
• B. $256$
• C. $2^9$
• D. $258$

Answer 1

The correct option is B $256$

$(1+x){}^9=$ $=1+{}^9{C}_{1}x+{}^9{C}_2{x}^2+{}^9{C}_3{x}^3+{}^9{C}_4{x}^4+{}^9{C}_5{x}^5+{}^9{C}_6{x}^6+{}^9{C}_7{x}^7+{}^9{C}_8{x}^8+{}^9{C}_9{x}^9$
Putting $x=1$, we get
$2^9=1+{}^9{C}_1+{}^9{C}_2+{}^9{C}_3+{}^9{C}_4+{}^9{C}_5+{}^9{C}_6+{}^9{C}_7{x}^7+{}^9{C}_8+{}^9{C}_9$ ...(i)
Putting $x=-1$, we get
$0=1-{}^9{C}_1+{}^9{C}_2-{}^9{C}_3+{}^9{C}_4-{}^9{C}_5+{}^9{C}_6-{}^9{C}_7{x}^7+{}^9{C}_8-{}^9{C}_9$ ...(ii)

Adding (i) and (ii), we get
$2^9=2[1+{}^9{C}_2+{}^9{C}_4+{}^9{C}_6+{}^9{C}_8]$
$2^8={}^9{C}_0+{}^9{C}_2+{}^9{C}_4+{}^9{C}_6+{}^9{C}_8$

Therefore,
${}^9{C}_0+{}^9{C}_2+{}^9{C}_4+{}^9{C}_6+{}^9{C}_8=2^8$ $=256$