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Algebra Level 3

1 2 + 2 2 + 3 2 + + 201 7 2 1 + 2 + 3 + + 2017 = ? \large \dfrac{1^2 + 2^2 + 3^2 + \cdots + 2017^2}{1+2+3+\cdots + 2017} = \, ?

1350 1349 1348 1345 1351 1347 1346 1352

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1 solution

Clearly the expression Can be simplified to - 2017 ( 2018 ) ( 4035 ) 6 2017.2018 2 = 4035 3 \frac{\frac{2017(2018)(4035)}{6}}{\frac{2017.2018}{2}} = \frac{4035}{3} . again don't divide it . we have only 1 option having unit's digit 5. So that's the answer indeed.

How can you simplify it to that?

Johnny Jillky - 5 years, 2 months ago

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The first expression is of the form :
1 2 + 2 2 + 3 2 + + n 2 1^2+2^2+3^2+\cdots+n^2 = n ( n + 1 ) ( 2 n + 1 ) 6 \frac{n(n+1)(2n+1)}{6} & the second one being 1 + 2 + 3 + + n = n ( n + 1 ) 2 1+2+3+\cdots +n=\frac{n(n+1)}{2} . Now put n= 2017 to get the desired result.

Aditya Narayan Sharma - 5 years, 2 months ago

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I see. Thanks!

Johnny Jillky - 5 years, 2 months ago

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