
Yefim S. answered 01/28/21
Math Tutor with Experience
Let (n + 1)th term is Tn + 1 = C(14, n)a14-n(b2)n = C(14, n)a14-nb2n. So, 2n = 6, n = 3 and we have to find 4th term of expansion: T4 = C(14, 3)a11b6 = (14·13·12)/(3·2·1) a11b6 = 364a11b6
Montse A.
asked 01/28/21The term containing b6
in the expansion of (a + b2)14
Yefim S. answered 01/28/21
Math Tutor with Experience
Let (n + 1)th term is Tn + 1 = C(14, n)a14-n(b2)n = C(14, n)a14-nb2n. So, 2n = 6, n = 3 and we have to find 4th term of expansion: T4 = C(14, 3)a11b6 = (14·13·12)/(3·2·1) a11b6 = 364a11b6
the 4th term is (a^12)^11(b^2)^3 which would give you a^11b^6
the coefficient of this term is 14 choose11(14C11)= 364
so the term is 364a^11b^6
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