Let $\alpha, \beta, \gamma$ and $\delta$ be the coefficients of $x^7, x^5, x^3$ and $x$ respectively in the expansion of $(x+\sqrt{x^3-1})^5 + (x-\sqrt{x^3-1})^5, x>1$. If $u$ and $v$ satisfy the equations $\alpha u+\beta v=18$ and $\gamma u+\delta v=20$, then $u+v$ equals:
- (1) 5
- (2) 4
- (3) 3
- (4) 8
Solution:
Expanding the term $(x+\sqrt{x^3-1})^5 + (x-\sqrt{x^3-1})^5$:
$$ = 2 \left[ ^5C_0 x^5 + ^5C_2 x^3(x^3-1) + ^5C_4 x(x^3-1)^2 \right] $$
$$ = 2 \left[ x^5 + 10x^3(x^3-1) + 5x(x^6-2x^3+1) \right] $$
$$ = 2 \left[ 5x^7 + 10x^6 + x^5 – 10x^4 – 10x^3 + 5x \right] $$
$$ = 10x^7 + 20x^6 + 2x^5 – 20x^4 – 20x^3 + 10x $$
Comparing coefficients: $\alpha = 10, \beta = 2, \gamma = -20, \delta = 10$.
Given equations:
1) $\alpha u + \beta v = 18 \Rightarrow 10u + 2v = 18 \Rightarrow 5u + v = 9$
2) $\gamma u + \delta v = 20 \Rightarrow -20u + 10v = 20 \Rightarrow -2u + v = 2$
Solving these, we get $u=1$ and $v=4$.
$$ \therefore u+v = 1+4 = 5 $$
Ans. (1)