Question

Coefficient of x^7 in the expansion of (1–2x+x^3)^5​

Answer 1

Answer:

(1+(x-x^2))^6

let X=x-(x^2)

according to binomial expansion formula,

1+ 6(x-x^2)+ (6*5/2!)*(x-x^2)^2+ (6*5*4/3!)*(x-x^2)^3+ (6*5*4*3/4!)*(x-x^2)^4+ (6*5*4*3*2/5!)*(x-x^2)^5+ (6*5*4*3*2*1/6!)*(x-x^2)^6

now in this expansion you wont find x^7 up to (x-x^2)^3 term, since highest power of x you will find is 6.

let, A=(6*5*4*3/4!)*(x-x^2)^4, B=(6*5*4*3*2/5!)*(x-x^2)^5, C=(6*5*4*3*2*1/6!)*(x-x^2)^6

in A, we can write (x-x^2)^4=(x^4)*(1+(-x))^4

in order to have x^7 term, we need to multiply (x^4) with x^3.

now if you expand (1+(-x))^4 , you will find the co-efficient of x^3 is -4.

so, (6*5*4*3/4!)*(x^4)*(-4*x^3)=-60*x^7

in B, we can write (x-x^2)^5=(x^5)*(1+(-x))^5

in order to have x^7 term, we need to multiply (x^5) with x^2.

now if you expand (1+(-x))^5 , you will find the co-efficient of x^2 is 10.

so, (6*5*4*3*2/5!)*(-10*x^7)=60*x^7

in C, we can write (x-x^2)^6=(x^6)*(1+(-x))^6

in order to have x^7 term, we need to multiply (x^6) with x.

now if you expand (1+(-x))^6 , you will find the co-efficient of x is -6.

so, (6*5*4*3*2*1/6!)*(-6*x^7)=6*x^7

finally (-60+60–6)*x^7=-6*x^7

and -6 should be your answer