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