Question

Find last digit of the number simplifying 1^2+2^2+3^2+.........+99^2

Answer 1

Answer:

Sum of the Squares of First n Positive Integers = n(n+1)(2n+1)6n(n+1)(2n+1)6

Putting n=99

Sum of the Squares of First 9999 Positive Integers = 99(100)(198+1)6=33∗50∗19999(100)(198+1)6=33∗50∗199

Last Digit of the number would be 0

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