Question

What is C(n, 1) + C(n, 2) + _ _ _ _  _ + C(n, n) equal to
1. 2 + 2² + 2³ + _ _ _ _ _  + 2ⁿ
2. 1 + 2 + 2² + 2³ + _ _ _ _ _ + 2ⁿ
3. 1 + 2 + 2² + 2³ + _ _ _ _ _ _ + 2^{n - 1}
4. 2 + 2² + 2³ + _ _ _ _ _ + 2^{n - 1}

Answer 1

Correct Answer - Option 3 : 1 + 2 + 2² + 2³ + _ _ _ _ _ _ + 2^{n - 1}

Concept:

(1 + x)n = nC0 × 1(n-0) × x 0+ nC1 × 1(n-1) × x 1 + nC2  × 1(n-2) × x2 + …. + nCn  × 1(n-n) × xn

n^th  term of the G.P. is an = arⁿ⁻¹

Sum of n terms = s = \(a (r^n-1)\over(r- 1)\); where r >1

Sum of n terms = s = \(a (1- r^n)\over(1- r)\); where r <1

Calculation:

C(n, 1) + C(n, 2) + _ _ _ _  _ + C(n, n) 

⇒ ⁿC₁ + ⁿC₂ + ... + ⁿC_n 

⇒ ⁿC₀ + nC1 + nC2 + ... + nCn - ⁿC₀

⇒ (1 + 1)ⁿ - ⁿC_o 

⇒ 2ⁿ - 1 = \(\rm 2^n - 1\over 2-1\) = 1 × \(\rm 2^n - 1\over 2-1\)

Comparing it with a G.P sum = a × \(\rm r^n - 1\over r-1\), we get a = 1 and r = 2

∴ 2n - 1 = 1 + 2 + 2² + ... +2^n-1 which will give us n terms in total.