If - represents the greatest integer function- then evaluate the following sum -11-12-22-13-23-33-14-24-34-44-15-upto the 2013th term- correct answer - 2- wrong answer 0

For 1≤ r lt;k,[rk]=0 The total number of terms upto [NN] is 12N(N+1), and we have the inequality 62×632=1953 lt;2013 lt;2016=63×642 So, the 2013^th term i ...

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Standard XII

Mathematics

Greatest Integer Function

If [.] repres...

Question

If $[.]$ represents the greatest integer function, then evaluate the following sum :
$[\frac{1}{1}] + [\frac{1}{2}] + [\frac{2}{2}] + [\frac{1}{3}] + [\frac{2}{3}] + [\frac{3}{3}] + [\frac{1}{4}] + [\frac{2}{4}] + [\frac{3}{4}] + [\frac{4}{4}] + [\frac{1}{5}] + \dots$
upto the $2013^{th}$ term.
(correct answer + 2, wrong answer 0)

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Solution

For $1 \leq r < k, [\frac{r}{k}] = 0$
The total number of terms upto $[\frac{N}{N}]$ is $\frac{1}{2} N (N + 1)$ , and we have the inequality $\frac{62 \times 63}{2} = 1953 < 2013 < 2016 = \frac{63 \times 64}{2}$
So, the $2013^{th}$ term is $[\frac{62}{63}]$
$∴$ Sum $= 1 + 1 + 1 + \dots 62 times = 62$

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If [.] represents the greatest integer function, then evaluate the following sum : [11]+[12]+[22]+[13]+[23]+[33]+[14]+[24]+[34]+[44]+[15]+⋯ upto the 2013th term. (correct answer + 2, wrong answer 0)

For 1≤r<k,[rk]=0 The total number of terms upto [NN] is 12N(N+1), and we have the inequality 62×632=1953<2013<2016=63×642 So, the 2013th term is [6263] ∴ Sum =1+1+1+⋯62 times=62