Question

Let n be an even integer and k =$\frac{3n}{2}$. Then the value of $\sum_{r = 1}^{k}{(–3)^{r–1}}$³ⁿC_{2r –1} is –

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

0

Answer 2

Since n is an even integer. Therefore, n = 2m, m Ī N.

Also, k = $\frac{3n}{2}$ Ž k = 3m.

\ $\sum_{r = 1}^{k}{(–3)^{r - 1}}$³ⁿC_2r–1

= $\sum_{r = 1}^{3m}{(–3)^{r - 1}}$^6mC_2r–1

= ^6mC₁ – ^6mC₃ (3) + ^6mC₅ (3)² – ^6mC₇ (3)³ + ….. + (–3)^{3m–1 6m}C_6m–1

Now, (1 + i$\sqrt{3}$)^6m = $\sum_{r = 0}^{6m}{6mC_{r}(i\sqrt{3})^{r}}$

Ž 2^6m $\left\{ \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} \right\}^{6m}$

= ^6mC₀ + ^6mC₁(i$\sqrt{3}$) + ^6mC₂ (i$\sqrt{3}$)² + ….+ ^6mC₃(i$\sqrt{3}$)³ + … + ^6mC_6m(i$\sqrt{3}$)^6m

Ž 2^6m (cos 2mp + i sin 2mp)

= {^6mC₀ – ^6mC₂ 3 + ^6mC₄ 3² – ….}

+i$\sqrt{3}${^6mC₁ – ^6mC₃ (3) + ^6mC₅(3)⁴ – ….}

[Using de moivre’s theorem on L.H.S.]

Equating imaginary parts on both sides, we get

2^6m sin 2mp = $\sqrt{3}${^6mC₁ – ^6mC₃ (3) + ^6mC₅ (3)⁴ – ….}

Ž ^6mC₁ – ^6mC₃ (3) + ^6mC₅ (3)⁴ – …. = 0 [Q sin 2mp = 0]

Ž $\sum_{r = 1}^{3m}{(–3)^{r - 1}}$^6mC_2r–1 = 0

Ž $\sum_{r = 1}^{k}{(–3)^{r - 1}}$³ⁿC_{2r – 1} = 0, where n = 2m and k = $\frac{3n}{2}$.

Hence (1) is correct answer.