The last digit of the number 1^{2} + 2^{2} + 3^{2} + ..... + 99^{2} is:

  • Option 1)

    0

  • Option 2)

    1

  • Option 3)

    6

  • Option 4)

    4

  • Option 5)

    3

Answers (1)

1^{2} + 2^{2} + 3^{2} + ..... + 99^{2}

\frac{n(n+1)(2n+1)}{6}

\frac{(99*100*199)}{6}

= 0

        

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