Let α, ß, γ and δ be the coefficients of x7, x5, x3 and x respectively in the expansion of x+x3−15+x−x3−15,x>1.If u and v satisfy the equationsαu+βv=18γu+δv=20then u + v equals :

x+x3−15+x−x3−15 =2 5C0⋅x5+5C2⋅x3x3−1+5C4⋅xx3−12 =25x7+10x6+x5−10x4−10x3+5x ⇒α=10,β=2,γ=−20,δ=10 Now, 10u+2v=18 −20u+10v=20 ⇒u=1,v=4 u+v=5

Let α, ß, γ and δ be the coefficients of x7, x5, x3 and x respectively in the expansion of x+x3−15+x−x3−15,x>1.If u and v satisfy the equationsαu+βv=18γu+δv=20then u + v equals :

x+x3−15+x−x3−15 =2 5C0⋅x5+5C2⋅x3x3−1+5C4⋅xx3−12 =25x7+10x6+x5−10x4−10x3+5x ⇒α=10,β=2,γ=−20,δ=10 Now, 10u+2v=18 −20u+10v=20 ⇒u=1,v=4 u+v=5

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Let α, ß, $γ$ and $δ$ be the coefficients of x⁷, x⁵, x³ 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
$\begin{aligned} αu + βv = 18 \\ γu + δv = 20 \end{aligned}$
then u + v equals :

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Detailed Solution

$\begin{array}{l} \begin{aligned} {(x + \sqrt{x^{3} - 1})}^{5} + {(x - \sqrt{x^{3} - 1})}^{5} \\ = 2 \{^{5} C_{0} \cdot x^{5} +^{5} C_{2} \cdot x^{3} (x^{3} - 1) +^{5} C_{4} \cdot x {(x^{3} - 1)}^{2}\} \\ = 2 \{5 x^{7} + 10 x^{6} + x^{5} - 10 x^{4} - 10 x^{3} + 5 x\} \\ \Rightarrow α = 10, β = 2, γ = - 20, δ = 10 \end{aligned} \\ Now, 10 u + 2 v = 18 \\ \begin{aligned} - 20 u + 10 v = 20 \\ \Rightarrow u = 1, v = 4 \\ u + v = 5 \end{aligned} \end{array}$