Calculator is Banal for Trivial Problems

Let, $201520142014=a$ .

Now, the expression becomes : $(a-1)^2-2a^2+(a+1)^2=2(a^2+1^2)-2a^2=\boxed2$

Let $201520142014 = x$

Thus, the given expression becomes,

$(x - 1)^2 - 2(x)^2 + (x+1)^2$

$(x^2 - 2x + 1) - 2x^2 +(x^2 + 2x +1)$

Simplifying leads to, every variable canceling out except $\boxed{2}$

All the numbers here have the numbers 2015201420 in common. You eliminate these numbers from each variable and you get the equation:

$13^{2} - 2(14)^{2} +15^{2} = 169-392+225=2$

$Let 201520142013 = a$

$a^{2} -2(a+1)^{2} + (a+2)^2$

$= a^{2} -2(a^{2}+2a+1+a^{2}+4a+4$

$= a^{2} -2a^{2}-4a-2+a^{2}+4a+4$

$= -2+4$

$= 2$

$\boxed{2}$

Got it by the fact that differences of squares are always odd, and if you order those differences they become a sequence of odd numbers: 4-1, 9-4, 16-7, ... is 3, 5, 7, ... You could write the expression as $201520142013^{2}$ - $201520142014^{2}$ + $201520142015^{2}$ - $201520142014^{2}$ The first two terms is the negative of their odd difference, while the last two makes the succeeding odd difference. The difference between two succeeding odd numbers is always two. Voila.

Consider 201520142013 as x, the equation becomes, a^2 - 2 (a+1)^2 + (a+2)^2, on expanding and solving, a^2 -2a^2 -4a-2+a^2+4a+4,which gives two(2)

Let 201520142013=a

So a²-2(a+1)²+(a+2)²

a²-2(a²+2a+1)+a²+4a+4

a²-2a²-4a-2+a²+4a+4

2a²-2a²-4a+4a+4-2 (so you organize and cancel the pairs)

2

$x=201520142013$ so
$x^2-2(x+1)^2+(x+2)^2=2$

Ending no. of those 3 numbers are 3,4,5 So now instead of using given values put 3,4,5 You will get 9-32+25 Which is = 2 Hence the answer!!!

Just look at the last digits since all the others are the same

3^2-2*(4^2)+5^2=34-32=2

Take the no 201520142014 as × then solve (×-1)^2 - 2×^2 +(× +1)^2 Then by the formula (a+b)^2 and (a-b)^2 solve it gives ×^2 -2×+1-2×^2+×^2+2×+1 It gives the answer 2

let

201520142015 = x 201520142014 = x-1 201520142013 = x-2

then,

(x-2)^2 -2(x-1)^2 +x^2 = 4-2 = 2 (ans)

becomes a^2-2(a+1)^+(a+2)^2 =a^2-2(a^2+2a+1)+4a^2+4a+4 =a^2-4a^2-4a-2+4a^2+4a+4 =2

If 201520142013 is x Then 201520142014 is x+1, and 201520142015 is x+2 So, x² - 2(x+1)² + (x+2)²
=x² - 2x² - 2 - 4x + x² + 4 +4x =4-2 =2

I took a longer route by knowing that the number in the middle is the average of the other two numbers

a=201520142013

b=201520142015

a^2 +-2 ((a+b)/2)^2+b^2=a^2-2 (a^2+2ab+b^2)/4+b^2=a^2/2-ab+b^2/2=(a-b)^2/2 and we know a-b=-2

then finally

(a-b)^2/2=2

Well this expression can be written as (321)^2 -2(322)^2+(323)^2 Note the pattern 123~201321042105

Let 201520142013 = n then 201520142014= n+1 and 201520142015= n+2 then the given equation becomes n^2 - 2 (n+1)^2 + (n+2)^2 = n^2 - 2n^2 - 4n - 2 +n^2 + 4n + 4 = 2

I didn't even think about it

20---------13=a, 20---------15=b,20-------14=(a+b)÷2,,,a²+b²-2(a+b)²÷4= (a-b)²÷2 a-b=2,,, 2²÷2=2

as a rule of squares: x"2 = (x-1)"2 + (x-1)*2 + 1

201520142013"2 -2(201520142014)"2 + 201520142015"2 = y

201520142013"2 - 201520142014"2 + 201520142014*2 +1 = y

201520142014 2 - 201520142013 2 = y

2(201520142014 - 201520142013) = y

y = 2

I did it algebraicly but you can do it in your head by substituting 1,2 and 3 in place of what was given

I started the solution in a similar way, let: $a = 201520142014$

Then we swap some elements in the expression: $(a-1)^{2}-2a^{2}+(a+1)^{2}=(a-1)^{2}+(a+1)^{2}-2a^{2}$

At this point I don't want to bother with multiplying the parenthesise out, so I do a little trick by adding and subtracting $2$ : $(a-1)^{2}+(a+1)^{2}-2a^{2}=(a-1)^{2}+(a+1)^{2}-2a^{2}+2-2$ $=(a-1)^{2}+(a+1)^{2}-2(a^{2}-1)-2$

Now we are able to reduce the expression by looking at this expression as a square: $(a-1)^{2}+(a+1)^{2}-2(a^{2}-1)-2=\big((a-1)-(a+1)\big)^{2}-2$

Now the $a$ 's cancel out and we get the answer: $\big((a-1)-(a+1)\big)^{2}-2=(a-a-1-1)^{2}-2=-2^{2}-2=4-2=2$