Question

Sol. [Key Concept: Integrate $(1 + x)^{12} = \sum{}^{12}C_{r}x^{r}$ term by term and evaluate between the given limits; the series equals $3^{13} - \alpha$ with $\alpha = 51$.]

$$(1 + x)^{12} = \ ^{12}C_{0} + \ ^{12}C_{1}x + \ ^{12}C_{2}x^{2} + \ldots\ldots.. + \ ^{12}C_{12}x^{12} $$integrating both sides

$$\left( \frac{(1 + x)^{13}}{13} \right)_{- 2}^{2} = \left( \ ^{12}C_{0}x + \ ^{12}C_{1}\frac{x^{2}}{2} + \ ^{12}C_{2}\frac{x^{3}}{3} + \ldots + \ ^{12}C_{12}\frac{x^{13}}{13} \right)_{- 2}^{2}$$

$$\frac{3^{13}}{13} + \frac{1}{13} = \left( \ ^{12}C_{0}2 + \ ^{12}C_{1}\frac{2^{2}}{2} + \ ^{12}C_{2}\frac{2^{3}}{3} + \ldots + \ ^{12}C_{12}\frac{2^{13}}{13} \right)$$

$$- \left( - \ ^{12}C_{0} \cdot 2 + \ ^{12}C_{1}\frac{2^{2}}{2} - \ ^{12}C_{2} \cdot \frac{2^{3}}{3} + \ldots + \ ^{12}C_{12} \cdot \frac{2^{13}}{13} \right)$$

$$= 2\left( 2.\ ^{12}C_{0} + \ ^{12}C_{2} \cdot \frac{2^{3}}{3} + \ ^{12}C_{4} \cdot \frac{2^{5}}{5} + \ldots + \ ^{12}C_{12} \cdot \frac{2^{13}}{13} \right)$$

$$3^{13} + 1 = 26\left( 2 + \ ^{12}C_{12} \cdot \frac{2^{3}}{3} + \ ^{12}C_{4} \cdot \frac{2^{5}}{5} + \ldots + \ ^{12}C_{12} \cdot \frac{2^{13}}{13} \right)$$

$$3^{13} - 51 = 26\left( \ ^{12}C_{2} \cdot \frac{2^{3}}{3} + \ ^{12}C_{4} \cdot \frac{2^{5}}{5} + \ldots + \ ^{12}C_{12} \cdot \frac{2^{13}}{13} \right)$$

$$\therefore\alpha = 51 $$