Let the coefficients of three consecutive terms T r , T r + 1 and T r + 2 in the binomial expansion of ( a + b ) 12 be in a G.P. and let p be the number of all possible values of r . Let q be th

Question

Let the coefficients of three consecutive terms T r ​ , T r + 1 ​ and T r + 2 ​ in the binomial expansion of ( a + b ) 12 be in a G.P. and let p be the number of all possible values of r . Let q be the sum of all rational terms in the binomial expansion of ( 4 3 ​ + 3 4 ​ ) 12 . Then p + q is equal to :

TestHub · Accepted Answer

Coefficient of ​ T r ​ , T r + 1 ​ , T r + 2 ​ → GP ⇒ ( 12 C r ​ ) 2 = 12 C r − 1 ​ ⋅ 12 C r + 1 ​ ​ ⇒ ( 12 C r ​ ) 2 = 12 C r − 1 ​ ⋅ 12 C r + 1 ​ but no three consecutive binomial coefficient are in GP ⇒ P = 0 Now for ( 3 1/4 + 4 1/3 ) 12 , T r + 1 ​ = 12 C r ​ ( 4 ) K /3 ( 3 ) 4 12 − K ​ for rational terms K = 0 , 12 sum of rational terms ​ = 12 C 0 ​ 4 0 ⋅ 3 3 + 12 C 12 ​ ⋅ 4 4 ⋅ 3 0 = 27 + 256 = 283 = q ∴ p + q = 283 ​

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