Some sums

Algebra Level 2

1 3 + 2 3 + + 201 6 3 ( 1 + 2 + + 2016 ) 2 = ? \large 1^3 + 2^3 + \cdots + 2016^3 - (1 + 2 + \cdots + 2016)^2 = \, ?


The answer is 0.00.

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3 solutions

Goh Choon Aik
Apr 14, 2016

tbh I guessed

This is not a solution - please refrain from posting in the solutions section if you have nothing to share.

Michael Fuller - 5 years, 2 months ago
Atul Shivam
Apr 12, 2016

Let 1 3 + 2 3 + 3 3 + 4 3 + . . . + 201 6 3 = ( 2016 × 2017 2 ) 2 . . . . . . . . . ( 1 ) 1^3+2^3+3^3+4^3+...+2016^3=(\frac{2016×2017}{2})^2.........(1)

Now ( 1 + 2 + 3 + 4 + . . . . + 2016 ) 2 = ( 2016 × 2017 2 ) 2 . . . . . . . . . ( 2 ) (1+2+3+4+....+2016)^2= (\frac{2016×2017}{2})^2.........(2) since both ( 1 ) , ( 2 ) (1),(2) are same so ( 1 ) ( 2 ) = 0 (1)-(2)=0

1 3 + 2 3 + + 201 6 3 ( 1 + 2 + + 2016 ) 2 1^3 + 2^3 + \cdots + 2016^3 - (1 + 2 + \cdots + 2016)^2

= ( 1 + 2 + + 2016 ) 2 ( 1 + 2 + + 2016 ) 2 =(1 + 2 + \cdots + 2016)^2-(1 + 2 + \cdots + 2016)^2

= 0 =\boxed0 .

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