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Find the last digit of the sum: 1^2 + 2^2 + 3^3 ....2017^2

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Find the last digit of the sum: 1^2 + 2^2 + 3^3 ....2017^2
posted Sep 25, 2017 by anonymous

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1 Answer

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Answer: 2737280785
the last digit of the sum = 5

Explanation:

1^2 + 2^2 + 3^3 ....2017^2
1+4+9+16+25+36+...+n^2=(n(n+1)(2*n+1))/6
1^2 + 2^2 + 3^3 ....2017^2 = (2017(2017+1)(2*2017+1))/6 = 2737280785
2737280785

answer Oct 5, 2017 by Mogadala Ramana



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