Let and be the coefficients of ands x respectively in the expansion of . If u and v satisfy the equations , then u + v equals : [2025]

Question

Let $α, β, γ$ and $δ$ be the coefficients of $x^{7}, x^{5}, x^{3}$ ands 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 $α u + β v = 18, γ u + δ v = 20$ , then u + v equals : [2025]

Smart Achievers · Accepted Answer

(1)

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

= $2 {{}^{5}C_{0} x^{5} + {}^{5}C_{2} x^{3} (x^{3} - 1) + {}^{5}C_{4} x {(x^{3} - 1)}^{2}}$

$∵ [{(x + a)}^{n} + {(x - a)}^{n} = 2 ({}^{n}C_{0} x^{n} + {}^{n}C_{2} x^{n - 2} a^{2} + . . .)]$

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

On comparing, we get

$α = 10, β = 2, γ = - 20, δ = 10$

Now, 10u + 2v = 18 and – 20u +10v = 20

$\Rightarrow$ u = 1, v = 4

Hence, u + v = 5.

Solution

Q.
Let $α, β, γ$ and $δ$ be the coefficients of $x^{7}, x^{5}, x^{3}$ ands 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 $α u + β v = 18, γ u + δ v = 20$ , then u + v equals : [2025]

1 5

2 3

3 4

4 8

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Solution

Q. Let α,β,γ and δ be the coefficients of x7,x5,x3 ands x respectively in the expansion of (x+x3–1)5+(x–x3–1)5, x>1. If u and v satisfy the equations αu+βv=18, γu+δv=20, then u + v equals : [2025]

1 5

2 3

3 4

4 8

Ans. (1) We have, (x+x3–1)5+(x–x3–1)5 = 2{C05x5+C25x3(x3–1)+C45x(x3–1)2} ∵ [(x+a)n+(x–a)n=2(C0nxn+C2nxn–2a2+...)] =2{5x7+10x6+x5–10x4–10x3+5x} On comparing, we get α=10, β=2, γ=–20, δ=10 Now, 10u + 2v = 18 and – 20u +10v = 20 ⇒ u = 1, v = 4 Hence, u + v = 5.

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Q.
Let $α, β, γ$ and $δ$ be the coefficients of $x^{7}, x^{5}, x^{3}$ ands 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 $α u + β v = 18, γ u + δ v = 20$ , then u + v equals : [2025]