Question

5. In the expansion, (1 + x) 14 , write the coefficient of x12

Answer 1

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Step-by-step explanation:

Let (r+1)th term contains x12 in the expansion of (1+x)14.Now, tr+1 = Cr14 . (1)14−r . xr .....(1)Put r = 12 in (1), we get t13 = C1214 . 1(14−12) . x12⇒ t13 = 14!2! . 12! × 12 . x12⇒t13 = 14 × 13 × 12!2! × 12! . x12⇒t13 = 14 × 132 ×x12⇒t13 = 91x12Coefficient of x12 = 91