 # Show that  11 2 + 12 2 +13 2 + .............+ 20 2 is an odd integer divisible by 5. AskIITians Expert Hari Shankar IITD
17 Points
14 years ago

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

14 years ago

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