Let , , and be the coefficients of x^7, x^5, x^3 and x respectively in the expansion of [a…

Question

Let $\alpha, \beta, \gamma$ and $\delta$ be the coefficients of $x^7, x^5, x^3$ and $x$ respectively in the expansion of

$$\begin{aligned}
& \left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5, x>1 \text {. If } u \text { and } v \text { satisfy the equations } \\
& \alpha u+\beta v=18, \\
& \gamma u+\delta v=20,
\end{aligned}$$

then $\mathrm{u+v}$ equals :

Accepted Answer

To find the sum of $ u $ and $ v $, we first need to expand the expression:

$ \left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5 $

Using the Binomial Theorem, the expansion yields:

$ = 2\left({}^5C_0 \cdot x^5 + {}^5C_2 \cdot x^3(x^3-1) + {}^5C_4 \cdot x(x^3-1)^2\right) $

Simplifying this, we obtain:

$ = 2\left(5x^7 + 10x^6 + x^5 - 10x^4 - 10x^3 + 5x\right) $

From this expansion, we can identify the coefficients:

The coefficient of $ x^7 $ is $ \alpha = 10 $

The coefficient of $ x^5 $ is $ \beta = 2 $

The coefficient of $ x^3 $ is $ \gamma = -20 $

The coefficient of $ x $ is $ \delta = 10 $

Given the equations:

$ \alpha u + \beta v = 18 $

$ \gamma u + \delta v = 20 $

Substituting in the coefficients:

$ 10u + 2v = 18 $

$ -20u + 10v = 20 $

By solving these equations, we find:

From $ 10u + 2v = 18 $, simplify to $ 5u + v = 9 $.

From $ -20u + 10v = 20 $, simplify to $ -2u + v = 2 $.

Solving these linear equations simultaneously, we find:

Subtracting equation 2 from equation 1:

$ 5u + v = 9  $

$ - (-2u + v = 2) $

This yields:

$ 7u = 7 \quad \Rightarrow \quad u = 1 $

Substitute $ u = 1 $ back into $ 5u + v = 9 $:

$ 5(1) + v = 9 \quad \Rightarrow \quad v = 4 $

Thus, the sum $ u + v $ is:

$ u + v = 1 + 4 = 5 $

Solutions · 1