Give the coefficient of x5 in the binominal expansion of (2x + 0.5)8.
A. 26,880
B. 224
C. 1,680
D. 1
E. 14
Correct Answer: B
Explanation/Reference:
Explanation:
If the expression (A x+ B)n is expanded, then by the binomial theorem, the xk term is C(n,k) × (Ax)k × Bn-k = C(n,k) × Ak × Bn-k × xk or, equivalently, the coefficient of xk is C(n,k) × Ak × Bn-k Therefore, the x5 coefficient can be determined by setting A = 2, B = 0.5, k = 5,n = 8:
C(8,5) × 25 × 0.58-5 = C(8,5) × 25 × 0.53 = 56 × 32 × 0.125= 224
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