Coefficient of Tr,Tr+1,Tr+2⟶GP ⇒(12Cr)2=12Cr−1⋅12Cr+1 but no three consecutive binomial coefficient are in GP ⇒P=0 Now for (31∕4+41∕3)12,Tr+1=12Cr(4)K∕3(3)
12−K
4
for rational terms K=0,12 sum of rational terms =12C040⋅33+12C12⋅44⋅30 =27+256=283=q ∴p+q=283
