Question

the constant term in the expression of (x+2/x)^6​

Answer 1

Answer:

Let y=x+2x

(1+x+2x)6=(1+y)6=∑k=06(6k)yk

For constant term, look for constant term in the expansion of yk

yk=(x+2x)k=∑r=0k(kr).xr.(2x)k−r

yk=∑r=0k(kr).xr−k+r.2k−r

yk=∑r=0k(kr).x2r−k.2k−r

∴(1+x+2x)6=∑k=06(6k)yk

=∑k=06(6k).∑r=0k(kr).x2r−k.2k−r

=∑k=06∑r=0k(6k).(kr).2k−r.x2r−k

Coefficient of any term is ∑k=06∑r=0k(6k).(kr).2k−r

For constant term: 2r−k=0

This is possible when:

k=0⟹r=0

C1=(60).(00).20−0=1

k=2⟹r=1

C2=(62).(21).22−1=

6∗51∗2∗2∗2=60

k=4⟹r=2

C3=(64).(42).24−2=

6∗5∗4∗31∗2∗3∗4∗4∗31∗2∗4=360

k=6⟹r=3

C4=(66).(63).26−3=

=1∗6∗5∗41∗2∗3∗8=160

Constant Term =C1+C2+C3+C4=1+60+360+160=581

Ans: 581