Q52 / 125 · BITSAT 2013 · Mathematics

The sum of the series $1+2.2+3.2^{2}+4.2^{3}+\ldots .+100.2^{99}$ is

  • A
    $100 \cdot 2^{100}+1$
  • Correct answer
    B
    $99.2^{100}+1$
  • C
    $99.2^{100}-1$
  • D
    $100.2^{100}+1$

Solution

Let $S=1+2.2+3.2^{2}+4.2^{2}+\ldots .+100.2^{99}$ It is an arithmetico - geometric series. On multiplying Eq. (i) by 2 and then subtracting it from Eq. (i), we get $S=1+2.2+3.2^{2}+4.2^{3}+\ldots+100.2^{99}$ $2 S=1.2+2.2^{2} \pm 3.2^{3}+\ldots+99.2^{99}+100.2^{100}$ $\begin{aligned}-S &=1+2+2^{2}+2^{3} \ldots+2^{99}-100.2^{100} \\ & \Rightarrow-S=\frac{1\left(2^{100}-1\right)}{2-1}-100.2^{100} \\ & \Rightarrow-S=2^{100}-1-100.2^{100} \\ & \Rightarrow-S=-1-99.2^{100} \\ & \Rightarrow S=99.2^{100}+1 \end{aligned}$

Hence, the correct answer is option: 'B'.

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