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BITSAT 2013
Exam
BITSAT
Year
2013
Subject
Mathematics
Chapter
Sequences and Series

BITSAT 2013 · Mathematics · Sequences and Series

The sum of the series 1+2.2+3.2^ 2 +4.2^ 3 + .+100.2^ 99 is

The sum of the series 1+2.2+3.22+4.23+….+100.2991+2.2+3.2^{2}+4.2^{3}+\ldots.+100.2^{99} is

Options

  1. A100⋅2100+1100 \cdot 2^{100}+1
  2. B99.2100+199.2^{100}+1
  3. C99.2100−199.2^{100}-1
  4. D100.2100+1100.2^{100}+1

Correct answer

B. 99.2100+199.2^{100}+1

Step-by-step solution

Let S=1+2.2+3.22+4.22+….+100.299S=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+3.22+4.23+…+100.2992S=1.2+2.22±3.23+…+99.299+100.2100S=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}−S=1+2+22+23…+299−100.2100⇒−S=1(2100−1)2−1−100.2100⇒−S=2100−1−100.2100⇒−S=−1−99.2100⇒S=99.2100+1\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}$$
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