What is 1-T_{1}+2T_{2}-3T_{3}+\dots+(-1)^{n}nT_{n} equal to?
Let (1+x)^{n}=1+T_{1}x+T_{2}x^{2}+T_{3}x^{3}+\dots+T_{n}x^{n}.
- A. 0
- B. -2^{n-1}
- C. n2^{n-1}
- D. 1 ✓
Correct Answer: D. 1
Explanation
The term sequence -T_1 + 2T_2 - 3T_3 \dots can be derived by evaluating the derivative n(1+x)^{n-1} = \sum_{k=1}^n k T_k x^{k-1} at x = -1. This evaluation yields 0 for n > 1. Therefore, the given expression simplifies to 1 + 0 = 1.
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