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

Question

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

$\begin{gathered} {(x+\sqrt{{x}^{3}-1})}^{5}+{(x-\sqrt{{x}^{3}-1})}^{5},x>1uv \ \alpha u+\beta v=18, \ \gamma u+\delta v=20, \end{gathered}$

then $u+v$ equals :

Accepted Answer

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

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

Using the Binomial Theorem, the expansion yields:

$=2({}^{5}{C}_{0}\cdot {x}^{5}+{}^{5}{C}_{2}\cdot {x}^{3}({x}^{3}-1)+{}^{5}{C}_{4}\cdot x({x}^{3}-1{)}^{2})$

Simplifying this, we obtain:

$=2(5{x}^{7}+10{x}^{6}+{x}^{5}-10{x}^{4}-10{x}^{3}+5x)$

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:

$7u=7\, \Rightarrow \, u=1$

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

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

Thus, the sum $u+v$ is:

$u+v=1+4=5$

Solutions · 1