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Algebra Level 3

What is the sum of the first 2016 2016 even numbers .

Important: The first even number will be 2 2 .

This problem is inspired with the "odd version of this", which is probably not original.


The answer is 4066272.

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3 solutions

\rightarrow Sum of first n even numbers = \text{Sum of first n even numbers}= n ( n + 1 ) n(n+1)
2016 × 2017 = 4066272 \Rightarrow 2016×2017=\boxed{4066272}

Sum of first n even numbers only n(n+1).You hasn't mentioned it.Please mention it

Vishal S - 5 years, 5 months ago

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Mentioned now!

A Former Brilliant Member - 5 years, 5 months ago
Ayushman Chahar
Jan 16, 2016

This can be done by using Arithmatic Progression

Milan Milanic
Jan 15, 2016

Solution:

Mathematical induction:

S ( 1 ) = 2 ; S(1) = 2;

S ( 2 ) = 2 + 4 = 6 ; S(2) = 2 + 4 = 6;

S ( 3 ) = 2 + 4 + 6 = 12 ; S(3) = 2 + 4 + 6 = 12;

. . . ...

It seems that S ( n ) = n ( n + 1 ) S(n) = n(n + 1) . If so, then: S ( n + 1 ) = 2 + 4 + . . . + 2 n + 2 n + 2 = n ( n + 1 ) + 2 ( n + 1 ) = S(n + 1) = 2 + 4 + ... + 2n + 2n + 2 = n(n + 1) + 2(n + 1) = = ( n + 2 ) ( n + 1 ) = (n + 2)(n + 1) .

Therefore, the solution is 2016 × 2017 = 4066272 2016 \times 2017 = \boxed{4066272} .

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