Let α and β be the roots of the equation x2-3 x+3=0. If S is the algebraic sum of the coefficients and T is the term independent of x in ((α+1) x +((β+1)/ x ))20, then

Question

Let α and β be the roots of the equation x2-3 x+3=0. If S is the algebraic sum of the coefficients and T is the term independent of x in ((α+1) x +((β+1)/ x ))20, then (A) S=520 (B) S =320 (C) T = 20 C 10 ⋅ 510 (D) T = 20 C 10 ⋅ 710

Tardigrade Admin · Accepted Answer

((α+1) x+((β+1)/x))20 S=(α+1+β+1)20=520 For the term independent of x T = 20 C r ⋅(α+1)20- r ⋅ x 20- r ⋅ ((β+1) r / x r ) 20-2 r =0 ⇒ r =10 ∴ T = 20 C 10 ⋅(α+1)10 ⋅(β+1)10= 20 C 10 ⋅(α β+α+β+1)10= 20 C 10 ⋅ 710

Q. Let $α$ and $β$ be the roots of the equation $x^{2} - 3 x + 3 = 0$ . If $S$ is the algebraic sum of the coefficients and $T$ is the term independent of $x$ in $((α + 1) x + \frac{( β + 1 )}{x})^{20}$ , then

$S = 5^{20}$

$S = 3^{20}$

$T =^{20} C_{10} \cdot 5^{10}$

$T =^{20} C_{10} \cdot 7^{10}$

Q. Let α and β be the roots of the equation x2−3x+3=0. If S is the algebraic sum of the coefficients and T is the term independent of x in ((α+1)x+x(β+1)​)20, then

S=520

S=320

T=20C10​⋅510

T=20C10​⋅710

((α+1)x+x(β+1)​)20 S=(α+1+β+1)20=520 For the term independent of x T=20Cr​⋅(α+1)20−r⋅x20−r⋅xr(β+1)r​ 20−2r=0⇒r=10 ∴T=20C10​⋅(α+1)10⋅(β+1)10=20C10​⋅(αβ+α+β+1)10=20C10​⋅710

Q. Let $α$ and $β$ be the roots of the equation $x^{2} - 3 x + 3 = 0$ . If $S$ is the algebraic sum of the coefficients and $T$ is the term independent of $x$ in $((α + 1) x + \frac{( β + 1 )}{x})^{20}$ , then

$S = 5^{20}$

$S = 3^{20}$

$T =^{20} C_{10} \cdot 5^{10}$

$T =^{20} C_{10} \cdot 7^{10}$