Question

Sum of coefficients of even powers of x in (1-x+x2-x3)^5 in binomial expansion example

Answer 1

hope that it is helpful to you

Answer 2

Answer:

Step-by-step explanation

{${1-x+x^{2}+x^{3} }$}^5

={${(1-x) + x^{2} (1-x^{2} })$}^5

=$(1-x)^5(1+x^{2} )^5$

=(1 - $^5C_{1}$x +$^5C_{2 }x^{2}$-$^5C_{3}x^{3}$+$^5C_{4}x^{4}$-$x^{5}$)(1 +$^5C_{1}x^{2}$+$^5C_{2}x^{4}$+$^5C_{3}x^{6}$+$^5C_{4}x^{8}$+$x^{10}$)

∴the sum of coefficients of even power of x is =$2^5C_{2}+2^5C_{4}+^5C_{1}+^5C_{3}+^5C_{5}$ +$( ^5C_{2}+^5C_{4})(1+^5C_{1}+ ^5C_{2}+^5C_{3}+^5C_{4}+^5C_{5})$ =20+10+5+10+1+(10+5)(1+5+10+10+5+1)=526