MY CART (5)

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

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping
Menu
Tushar Watts Grade: 12
        

Show that  11 2 + 12 2 +13 2 + .............+ 20 2 is an odd integer divisible by 5.

7 years ago

Answers : (2)

AskIITians Expert Hari Shankar IITD
17 Points
										

Sum of squares of first n consecutive integers = n(n+1)(2n+1)/6.


Therefore, 12 +22 + ...+192 +202 = 20 * 21 * 41 /6 = 2870.............(1)


Also, 12 + 22 + 32 + ...+ 102 = 10*11*21/6 = 385...................(2)


 


Subtracting (2) from (1),


the first 10 terms of (1) get cancelled out and we get


112 + 122 + 132 = ....+ 192 +202 = 2870-385 = 2485, which is an odd integer and is divisible by 5


 


 

7 years ago
Ramesh V
70 Points
										

This is simply:


S= sum ( 12+22+32+...202) - sum (12+22+32+...102)


as (12+22+32+........+n2 ) = n(n+1)(2n+1)/6


S = 2485 which is divisible by 5..


If this answer is wat u r not expecting, plzz post question properly ..


--


Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and

we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best.



Regards,

Naga Ramesh

IIT Kgp - 2005 batch

7 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: R 15,000
  • View Details
Get extra R 3,000 off
USE CODE: CART20
Get extra R 440 off
USE CODE: CART20

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details