Binomial Theorem Part 1

Let r=2
∴2‌nd, 3rd and 4th terms are in AP.
2nd term =T2=‌mC1⋅y
Coefficient of T2=‌mC1
3rd term =T3=‌mC2⋅y2
Coefficient of T3=‌mC2
4th term =T4=‌mC3⋅y3
Coefficient of T2=‌mC3
∴2.‌mC2=‌mC1+‌mC3
⇒2.‌

m(m−1)

1.2

=‌

m

1

+‌

m(m−1)(m−2)

1.2.3

⇒6m2−6m=6m+m(m2−3m+2)
⇒6m2−6m=6m+m3−3m2+2m
⇒6m−6=6+m2−3m+2
⇒m2−9m+14=0
Now put r=2 at each option and find answer.
In option C, m2−m(4r+1)+4r2−2=0 putting r=2 we get m2−9m+14=0. So Option C is correct.