What is T_{1}+T_{2}+T_{3}+\dots+T_{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. 2^{n}
  2. B. 2^{n}-1 ✓
  3. C. 2^{n-1}
  4. D. 2^{n}+1

Correct Answer: B. 2^{n}-1

Explanation

Substituting x=1 into the identity (1+x)^n = 1 + T_1 x + T_2 x^2 + \dots + T_n x^n gives 2^n = 1 + T_1 + T_2 + \dots + T_n. Subtracting 1 from both sides gives T_1 + T_2 + \dots + T_n = 2^n - 1.

Related questions on Algebra

Practice more NDA Mathematics questions