1. Tardigrade
  2. Question
  3. Mathematics
  4. If (x-2)100= displaystyle∑r=0100 ar ⋅ xr, then a1+2 a2+ ldots+100 a100=

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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

Binomial Theorem

Solution:

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