vdsoccer Wrote:What will be the tens digit of :
99^2 + 999^2 + 9999^2........................999999999999^2
split the terms
99^2 + 999^2 +..........
as
(100 - 1)^2 + (1000 - 1)^2 + ........
now (100 - 1)^2 = 10000 - 200 + 1
(1000 - 1)^2 = 1000000 - 2000 + 1
and so on. You don't have to calculate them all, just keep in mind that each one of them has a 0 at ten's digit and 1 at the unit's digit. Thus you get 11 such terms. add them up and the last two digits of the sum will be 11 thus the answer to your question is 1
You can use the same approach for other such series as
98^2 + 998^2 + .......
or
101^2 + 1001^2 + .......