What is T_{1}+2T_{2}+3T_{3}+\dots+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}.

  1. A. 0
  2. B. 1
  3. C. 2^{n}
  4. D. n2^{n-1}

Correct Answer: D. n2^{n-1}

Explanation

Differentiating both sides of (1+x)^n = 1 + T_1 x + T_2 x^2 + \dots + T_n x^n with respect to x yields n(1+x)^{n-1} = T_1 + 2T_2 x + 3T_3 x^2 + \dots + nT_n x^{n-1}. Substituting x=1 provides the required sum: n(2)^{n-1}.

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