2014 Frenzy with Squares

Applying the identity: $a^2-b^2=(a+b)(a-b)$

$2014^2-2013^2+2012^2-2011^2...2^2-1^1=(2014+2013)+(2012+2011)+(2010+2009)...+(2+1)$

$= \frac{2014 \times 2015 }{2}=2029105$

So the answer is $\boxed{105}$

The sum can be written like this form $\sum_{n=1}^{1007}(2n)^2 - \sum_{n=1}^{1007}(2n-1)^2,$ squaring both sides and using properties of the sum we have $\sum_{n=1}^{1007}4n^2 - (4n^2-4n+1)=\sum_{n=1}^{1007}4n-1,$ using properties of the sum one more time $4\sum_{n=1}^{1007}n-\sum_{n=1}^{1007}1,$ using the formulas $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}\\\sum_{k=1}^{n}1=n,$ so $4\frac{(1007)(1008)}{2}-1007=4043105,$ then the last three digits are $\boxed{105}$

Here, if from the start, we evaluate consecutive two terms of the series, we get an AP as: $4027,4023,....3$ .

Let us take $a , l , d , n$ as the first term, last term, common difference and no. of terms of the AP resp.

Let $a_{n}=3 \implies a+(n-1)d=3 \implies 4027+(n-1)(-4)=3 \implies 4(n-1)=4024 \implies n-1=1006 \implies n=1007$

So, there are 1007 terms in the AP as 3 is the last term.

Now, Sum $=\frac{n}{2} (a+l) = \frac{1007}{2} (3+4027) = \frac{1007}{2}\times 4030= (1007\times 2015) = \boxed{2029105}$

Here, we have the last 3 digits of the sum as $\boxed{105}$ .

taking two terms and solving ,we get an A.P. i.e. 4027 + 4023 + 4019 + .......................+3 , now , calculate no. of terms by p = a + ( n-1 )d , here , p =3 ,a=4027 , d = (-4) , solving we get n=1007 , now , sum of series = n [ 2a +(n-1) d] /2 , putting values we get sum =2029105 , so , last digits are 105

Note that $2^{2} - 1^{2} = 3$ , $4^{2} - 3^{2} = 7$ , $6^{2} - 5^{2} = 11$

Generally: $(2k)^{2} - (2k - 1)^{2} = 3 + 4(k - 1) = 4k - 1$ .

So we can evaluate the last one: $4(1007) - 1 = 4027$ .

Now, we can sum that with the first one ( $3$ ), and we get $4030$ . See that this will be repeating $503$ times until you get to $1008^{2} - 1007^{2}$ . Evaluating it similarly, as above, it is equal to $2015$ . In conclusion, our answer is $4030\cdot503 + 2015 = 2029105$ , and so our answer is $\boxed {105}$ .

$2014^{2}-2013^{2}+2012^{2}-2011^{2}+.....+2^{2}-1^{2}$ $=(2014+2013)(2014-2013)+(2012+2011)(2012-2011)+...+(2+1)(2-1)$ $=(2014+2013)\times 1+(2012+2011)\times 1+...+(2+1)\times 1$ $=2014+2013+2012+2011+...+2+1=\frac{2014\times 2015}{2}=2029105$ so, 105

2²-1=3= a1 ; 4²-3²=7= a2 ; 6²-5²=11= a3 ; 8²-7²=15 = a4 ; ... assim há uma P.A (progressão aritmética) de razão 4 onde o an pode ser escrito com [(2n)² - (2n-1)²] com n variando de 1 a 1007 ou uma P.A de razão 4 cujo a1 = 3 e possua 1007 termos. Assim recorrendo a soma de P.A S=(a1+ an)*n/2 = (3+4027)1007/2= 2.029.105

Taking the whole series by pairs of two, we get the required sum to be:3+7+11+....+4027 This sums up to (2015)(1007) = 2029105.

$2014^2-2013^2+2012^2-2011^2+...+2^2-1^2$ $= (2014-2013)(2014+2013)+(2012-2011)(2012+2011)+...+(2-1)(2+1)$ $=2014+2013+2012+2011+...+2+1$ $=\frac{(1+2014)*2014}{2}$ $=2029105$ get the last three digits, we will have the answer $\boxed{105}$

First you realize that this is murica, the land of the free and the home of the brave

Second, you realize 4-1 = 3 16-9 = 7 36-25 = 11

and so on..

Then using these murican powers, you realize using real eyes that:

This is just (1007) * (3+2012)

2014^2−2013^2 = (2014+2013)(2014-2013) = 2014+2013 2012^2−2011^2 = (2012+2011)(2012-2011) = 2012+2011 So, we have 2014 + 2013 + 2012 + 2011 + 2010 + 2009 + 2008 +...+ 2 + 1. 2014 + 1 = 2015; 2013 + 2 = 2015... and 2014/2 = 1007. So, 2014 + 2013 + 2012 + 2011 + 2010 + 2009 + 2008 +...+ 2 + 1 = 2015 x 1007 = 2.029.105. The result is 105

Each pair is a^2 - (a-1)^2 = a + a-1

Series = 2014 + 2013 + 2012 + 2011 + ... + 1 = 2014 * 2015 / 2

Sum of n = n/2(a=l) = 2014/2(2014+1) = 1007(2015) = 2029105 So the last three digits are 1,0,5 =105

(2014^2 - 2013^2)+(2012^2 - 2011^2) ...........+(2^2 -1^2)
=(2014+2013)(2014-2013)+(2012+2011)(2012-2011)..................+(2 +1)(2 -1) =2014+2013+2012+2011+...........2 +1 =(2014*2015)/2
=2029105

So, the last 3 digits are 105

According to $a^2 - b^2 = (a + b)(a - b)$

$(2014 + 2013)(2014 - 2013) + (2012 + 2011)(2012 - 2011) + \dots + (2 + 1)(2 - 1)$ $= 4027 + 4023 + \dots + 3$

There are 1007 terms,

So, $n = 1007$

$a_1 = 4027$

$a_n = 3$

Using $S = \frac {n}{2}(a_1 + a_n)$

$S=\frac {1007}{2}(4027 + 3)$

$S=2029105$

$2029 \boxed{105}$

So, the answer is $105$

(2014+2013)(2014-2013)+(2012+2011)(2012+2011)+........................+(2+1)(2-1) =1+2+3+4+5..........................+2011+2012+2013+2014=1021105

Using the algebraic identity

$a^2-b^2=(a+b)(a-b)$

we can rewrite our expression as

$(2014+2013)(2014-2013)+(2012+2011)(2012-2011)+..........+(2+1)(2-1)$

which is just $2014+2013+2012+2011+............+3+2+1$ .The sum is just

$\dfrac{2014\cdot 2015}{2}=1007\cdot 2015$

Since we only want the last three digits,we look at the final expression $\text{mod}$ $1000$ and that gives us 105.

Here we apply the formula a's power 2-b's power 2=(a-b)(a+b) so 2014 power 2- 2013 power 2=(2014-2013)(2014+2013) it will become 1*(2014+2013)=2014+2013
finally we found a series: 2014+2013+. . . +2+1 and solve this with the formula n(n+1)/2 so 2014(2014+1)/2=2029105 and last three digits are 105

it can be write like this

(2014 - 2013) (2014 + 2013) + (2012 - 2011)(2012+2011) + ... + (2-1)(2+1)

it form an aritmethic formula

Sn = $\frac{1}{2}$ n (2a + (n - 1) b )

Sn = $\frac{1}{2}$ 2014 (2 . 1 + (2014-1) 1)

Sn = 1007 $\times$ 2015

Sn = 2029105

the last three digits of the alternating sum is $\boxed{105}$

The problem above can change into (2014+2013)(2014-2013) + (2012+2011)(2012-2011) + ... + (4+3)(4-3) + (2+1)(2-1) = (2014+2013).1 + (2012+2011).1 + ... + (2+1).1 = 1 + 2 + 3 + ... + 2014 = 2014.2014/2 = ......105. Answer : 105

*we can rewrite the expression in the following way: *

$(2014^{2} - 2013^{2}) + (2012^{2} - 2011^{2}) + ................+(2^{2} - 1^{2})$

• by applying the formula* $a^{2} - b^{2} = (a+b)*(a -b)$ *,the expression becomes: *

$(2014 - 2013)*(2014 + 2013) + (2012 - 2011)*(2012 + 2011) +................+(2 - 1)*(2 + 1)$

• the final expression becomes: *

$4027 + 4023 +.........+ 3$

this is nothing but an Arithmetic Progression of first term 4027 , $n^{th}$ term 3 and common difference -4

• the value of n can be found out in the following manner*

$3 = 4027 - ( n -1)* 4$

equating this we get $n = 1007$

sum of 1007 terms S = $\frac{1007}{2}*( 4027 + 3)$ $= ...........105$

so the last three digits of the sum are 1 0 and 5

It can be written as 2014+2013+2012...............+1