If (x-2)100= displaystyle∑r=0100 ar ⋅ xr, then a1+2 a2+ ldots+100 a100=

Question

If (x-2)100= displaystyle∑r=0100 ar ⋅ xr, then a1+2 a2+ ldots+100 a100= (A) 100 (B) -100 (C) 1 (D) 101

Tardigrade Admin · Accepted Answer

(x-2)100= displaystyle∑r=0100 ar ⋅ xr Differentiating both sides w.r.t. x, we get 100(x-2)99= displaystyle∑r=0100 r . ar ⋅ xr-1 Putting x=1, we get 100(1-2)99=a1+2 a2+ ldots+100 a100 ∴ a1+2 a2+ ldots+100 a100=-100

Q. If $(x - 2)^{100} = r = 0 \sum 100 a_{r} \cdot x^{r},$ then $a_{1} + 2 a_{2} + \dots + 100 a_{100} =$

100

-100

1

101

$(x - 2)^{100} = r = 0 \sum 100 a_{r} \cdot x^{r}$
Differentiating both sides w.r.t. $x,$ we get
$100 (x - 2)^{99} = r = 0 \sum 100 r . a_{r} \cdot x^{r - 1}$
Putting $x = 1,$ we get
$100 (1 - 2)^{99} = a_{1} + 2 a_{2} + \dots + 100 a_{100}$
$∴ a_{1} + 2 a_{2} + \dots + 100 a_{100} = - 100$

Q. If (x−2)100=r=0∑100​ar​⋅xr, then a1​+2a2​+…+100a100​=

100

-100

1

101

(x−2)100=r=0∑100​ar​⋅xr Differentiating both sides w.r.t. x, we get 100(x−2)99=r=0∑100​r.ar​⋅xr−1 Putting x=1, we get 100(1−2)99=a1​+2a2​+…+100a100​ ∴a1​+2a2​+…+100a100​=−100

Q. If $(x - 2)^{100} = r = 0 \sum 100 a_{r} \cdot x^{r},$ then $a_{1} + 2 a_{2} + \dots + 100 a_{100} =$

$(x - 2)^{100} = r = 0 \sum 100 a_{r} \cdot x^{r}$
Differentiating both sides w.r.t. $x,$ we get
$100 (x - 2)^{99} = r = 0 \sum 100 r . a_{r} \cdot x^{r - 1}$
Putting $x = 1,$ we get
$100 (1 - 2)^{99} = a_{1} + 2 a_{2} + \dots + 100 a_{100}$
$∴ a_{1} + 2 a_{2} + \dots + 100 a_{100} = - 100$