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  4. The sum to 10 terms of the series 1+2(1.1)+3(1.1)2+4(1.1)3+.... is

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Q. The sum to $10$ terms of the series $1+2\left(1.1\right)+3\left(1.1\right)^{2}+4\left(1.1\right)^{3}+....$ is

NTA AbhyasNTA Abhyas 2022

Solution:

Let $x=1.1$
$S=1+2x+3x^{2}+........10x^{9}$ and
$xS=x+2x^{2}+........9x^{9}+10x^{10}$
Subtracting, we get,
$S\left(1 - x\right)=\left(1 + x + x^{2} + . . . . . . . + x^{9}\right)-10x^{10}$
$=\left(\frac{x^{10} - 1}{x - 1}\right)-10x^{10}$
$\Rightarrow -S=\frac{x^{10} - 1}{\left(x - 1\right)^{2}}-\frac{10 x^{10}}{\left(x - 1\right)}$
$\Rightarrow -S=\frac{\left(1.1\right)^{10} - 1}{\left(0.1\right)^{2}}-\frac{10 \left(1.1\right)^{10}}{0.1}=100\times \left(1.1\right)^{10}-100-100\times \left(1.1\right)^{10}=-100$
$\Rightarrow S=100.$