If [ x ] is the greatest integral function, then ∑ k = 1 4020 [ 2 1 + 4020 k − 1 ] is equal to

For k = 1 , 2 , 3 …… upto 2010, the value of [ 2 1 + 4020 k − 1 ] is equal to zero For k = 2011 , 2010 , ---- 4020, the value of [ 2 1 + 4020 k − 1 ] = 1 ∴ The sum value is 2010.

Start Free Mock LoginAlready registered? Login here

Mathematics - Functions Question with Solution | TestHub

MathematicsFunctionsGeneral (Definition,type and special type)Easy2 minQB

MathematicsEasysingle choice

If $[x]$ is the greatest integral function, then $\sum_{k = 1}^{4020} [\frac{1}{2} + \frac{k - 1}{4020}]$ is equal to

Options:

Answer:

Solution:

For $k = 1, 2, 3 \dots\dots$ upto 2010, the value of $[\frac{1}{2} + \frac{k - 1}{4020}]$ is equal to zero For $k = 2011, 2010$ , ---- 4020, the value of $[\frac{1}{2} + \frac{k - 1}{4020}] = 1$

∴ The sum value is 2010.

Stream:JEESubject:MathematicsTopic:FunctionsSubtopic:General (Definition,type and special type)

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...