Question

The sum to $50$ terms of the series $1+2\left(\frac{11}{50}\right)+3{\left(\frac{11}{50}\right)}^2+\dots$ is given by

• A. $2500$
• B. $2550$
• C. $2450$
• D. none of these

Answer 1

The correct option is D none of these

$S_{50}=1+2\left(\frac{11}{50}\right)+3{\left(\frac{11}{50}\right)}^2+\cdots +50{\left(\frac{11}{50}\right)}^{49}$
$\frac{11}{50}{S}_{50}=\frac{11}{50}+2{\left(\frac{11}{50}\right)}^2+3{\left(\frac{11}{50}\right)}^3+\cdots +50{\left(\frac{11}{50}\right)}^{50}$
$S_{50}-\frac{11}{50}{S}_{50}=1+\frac{11}{50}+{\left(\frac{11}{50}\right)}^2+{\left(\frac{11}{50}\right)}^3+\cdots +{\left(\frac{11}{50}\right)}^{49}-50{\left(\frac{11}{50}\right)}^{50}$
$\Rightarrow \frac{39}{50}{S}_{50}=\frac{1-{\left(\frac{11}{50}\right)}^{50}}{1-\frac{11}{50}}-50{\left(\frac{11}{50}\right)}^{50}$
$\Rightarrow S_{50}={\left(\frac{50}{39}\right)}^2[1-{\left(\frac{11}{50}\right)}^{50}]-\frac{{50}^2}{39}{\left(\frac{11}{50}\right)}^{50}$