0

Guest

Courses New

[sqrt(1)]+[sqrt(2)]+[sqrt(3)]+.......+[sqrt(2115)]=?

[sqrt(1)]+[sqrt(2)]+[sqrt(3)]+.......+[sqrt(2115)]=?

Grade:10

1 Answers

Arun Kumar IIT Delhi
askIITians Faculty 256 Points
8 years ago
Hello Student,

\\\left \lfloor \sqrt1 \right \rfloor=1 \\\left \lfloor \sqrt2 \right \rfloor=1 \\=>n=1=>1,2,3=>(2^2-1^2) \\=>n=2=>4,5,6,7,8=>(3^2-2^2) \\\sqrt{2014}=44 \\=>\sum_{1}^{43}((r+1)^2-r^2)=\sum_{1}^{43}(2r+1)=n(n+1)+n=43*44+44 \\$after 44 starts $add=44*(2014-1936+1)
Thanks & Regards
Arun Kumar
Btech, IIT Delhi
Askiitians Faculty

Think You Can Provide A Better Answer ?

Other Related Questions on Magical Mathematics[Interesting Approach]

View all Questions »

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More