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

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

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..

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