Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

[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
6 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 ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free