Sequences and Series 2 Question 3
3. For $x \in R$, let $[x]$ denote the greatest integer $\leq x$, then the sum of the series $-\frac{1}{3}+-\frac{1}{3}-\frac{1}{100}+-\frac{1}{3}-\frac{2}{100}+\ldots+-\frac{1}{3}-\frac{99}{100}$ is (2019 Main, 12 April I)
(a) $\frac{1}{m n}$
(b) $\frac{1}{m}+\frac{1}{n}$
(c) 1
(d) 0
Analytical and Descriptive Question
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Answer:
Correct Answer: 3. (c)
Solution:
- Given series is
$$ -\frac{1}{3}+-\frac{1}{3}-\frac{1}{100}+-\frac{1}{3}-\frac{2}{100}+\ldots \ldots+-\frac{1}{3}-\frac{99}{100} $$
[where, $[x]$ denotes the greatest integer $\leq x]$
Now,
$$ -\frac{1}{3},-\frac{1}{3}-\frac{1}{100},-\frac{1}{3}-\frac{2}{100}, \ldots+-\frac{1}{3}-\frac{66}{100} $$
all the term have value -1
and $-\frac{1}{3}-\frac{67}{100},-\frac{1}{3}-\frac{68}{100}, \ldots,-\frac{1}{3}-\frac{99}{100}$ all the term have value -2 .
So, $-\frac{1}{3}+-\frac{1}{3}-\frac{1}{100}+-\frac{1}{3}-\frac{2}{100}+\ldots+-\frac{1}{3}-\frac{66}{100}$
$=-1-1-1-1 \ldots 67$ times.
$=(-1) \times 67=-67$
and $-\frac{1}{3}-\frac{67}{100}+-\frac{1}{3}-\frac{68}{100}+\ldots+-\frac{1}{3}-\frac{99}{100}$
$=-2-2-2-2 \ldots 33$ times
$=(-2) \times 33=-66$ $\therefore \quad-\frac{1}{3}+-\frac{1}{3}-\frac{1}{100}+-\frac{1}{3}-\frac{2}{100}+\ldots+-\frac{1}{3}-\frac{99}{100}$
$=(-67)+(-66)=-133$.
Alternate Solution
$\because[-x]=-[x]-1$, if $x \notin$ Integer,
and $[x]+x+\frac{1}{n}+x+\frac{2}{n}+\ldots+\quad x+\frac{n-1}{n}=[n x]$,
$n \in N$.
So given series
$$ \begin{aligned} & -\frac{1}{3}+-\frac{1}{3}-\frac{1}{100}+-\frac{1}{3}-\frac{2}{100}+\ldots \ldots+\frac{-1}{3}-\frac{99}{100} \\ & =-\frac{1}{3}-1+-\frac{1}{3}+\frac{1}{100}-1 \\ & \quad+-\frac{1}{3}+\frac{2}{100}-1+\ldots+-\frac{1}{3}+\frac{99}{100}-1 \\ & =(-1) \times 100-\frac{1}{3} \times 100=-100-33=-133 . \end{aligned} $$
