Question

Find the coefficient of x in the
expansion of 122- elo​

Answer 1

Answer:

(1+x2)(x/2−4/x)6=T1∗T2

Binomial expansion of T2=(x/2−4/x)6=

∑r=06(6r)(−1)r(x/2)r(4/x)6−r

=∑r=06(6r)(−1)r2−r46−rxr−(6−r)

=∑r=06(6r)(−1)r2−r22(6−r)xr−(6−r)

=∑r=06(6r)(−1)r2−r+12–2rx2r−6

=∑r=06(6r)(−1)r212–3rx2r−6

Interested in the coefficient of x2

This can happen in 2 ways:

constant term of T1 is multiplied by the coefficient of x2 of T2

=>2r−6=2=>r=4

Constant of T1=1

Coefficient of x2 of T2 :

Substituting r=4 in (6r)(−1)r212–3r

(64)(−1)4212–3(4)=6/1∗5/2∗1∗2°=15

Coefficient of x2 = Constant of T1 * Coefficient of x2 of T2 =C1=1∗15=15

x2 of T1 is multiplied by constant term of T2

Coefficient of x2 of T1=1

Constant term of T2 = when x is raised to the power of 0

=>2r−6=0=>r=3

Substituting r=3 in (6r)(−1)r212–3r

(63)(−1)3212–3(3)=6/1*5/2∗4/3∗(−1)∗23=−160

Coefficient of x2 = Coefficient of x2 of T1 * Constant of T2=C2=1∗(−160)=−160

Coefficient of x2 in the expanded form =C1+C2=15−160=−145

Ans: -145