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) $-153$
(b) $-133$
(c) $-131$
(d) $-135$
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Answer:
Correct Answer: 3. (b)
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} $
