More even than odd

We have to find (2+4+6+.......98+100)-(1+3+5+.....97+99) =(2-1)+(4-3)+(6-5)+......(99-97)+(100-99) =1+1+1+1+......1+1 upto 50 times =50

Even - Odd = 1 Example: 2 - 1 = 1; 4 - 3= 1 and so on From 1-100 there are 50 pairs of odd and even numbers having the difference of 1 Therefore: 1x50=50

Since the answer is:

$(2+4+6+...+98+100)-(1+3+5+...+97+99)$

$2+4+6+...+98+100-1-3-5-...-97-99$

We could group them into:

$(2-1)+(4-3)+(6-5)+...+(98-97)+(100-99)$

$=1+1+1+...+1+1$ (50 times)

$=1 \times 50=50$

1+2+3+...+n=(n(n+1))/2

1+2+3+...+100=(100(100+1))/2=5050

1+3+5+...+(2n-1)=n^2

1+3+5+...+99=50^2=2500

2+4+6+...+100=5050-2500=2550

(2+4+6+...+100)-(1+3+5+...+99)=2550-2500=50

Take the first and last: 2 + 100 = 102 Then take the second and the second-to-last: 4 + 98 = 102 Then take the third and the third-to-last: 6 + 96 = 102

So you can see that we can get a bunch of pairs, all adding up to 102. How many pairs? Well, there are 50 even numbers less than or equal to 100, so there must be 25 pairs.

So Therefore 25(102) = 2550.

Take the first and last: 1 + 99 = 100 Then take the second and the second-to-last: 3 + 97 = 100 Then take the third and the third-to-last: 5 + 95 = 100

all adding up to 100. How many pairs? Well, there are still 50 even numbers less than or equal to 100, so there must be 25 pairs same as even have.

So Therefore 25(100) = 2500.

SO THE ANSWER IS 2550-2500 = "50"

On close observation we find that if we take the sum of first 2 odd and even positive integers, their difference is 2. Similarly if we take the sum of first 3 odd and even positive integers, their difference is 3. So if there are 50 odd and even positive integers between 0 to 100, their cumulative difference is 50.

Worded Explanation: Every Positive Odd Integer starting at 1, is one less than its corresponding Positive Even Integer. For Example, the Corresponding Even Integer of 1 is 2. Therefore, every Positive Odd Integer that follows 1 will always be 1 less than its Corresponding Even Integer. In 1-100 inclusive, there are exactly 50 Odd and 50 Even Integers. As a result, The Sum Of All Odd Integers will be 50 less than The Sum Of All Even Integers. (Extra:- Therefore since 50 times 1 is 50), the difference between the Sum of All Odd Integers between 1-100 inclusive and the Sum of All Even Integers between 1-100 inclusive is 50.

Mathematical Explanation: Arithmetic Progression

Let 1.) Be all Even integers Between 1 and 100, Inclusive 2.) Be all Odd Integers Between 1 and 100, Inclusive 3.) x = 50, The difference between the Sum of All Even and Odd Integers between 1-100 inclusive. 4.) sub. stands for subscript.

x(e sub.1 + e sub.x)/2 which results to 50(2+100)/2 = 50(102)/2 = 5100/2 = 2550. *2550 is the Sum Of All Even Integers Between 1-100 inclusive.

x (o sub.1+ o sub.2)/2 which result to 50(1+99)/2 = 50(100)/2 = 5000/2 =2500 *2500 is the Sum Of All Odd Integers Between 1-100 Inclusive.

Therefore, the difference between the Sum Of All Even and Odd Integers Between 1-100 Inclusive is (2550-2500), which equals 50.

Answer: 50

sum of natural no=n(n+1)/2............(1) sum of all odd no=n^2..............(2) where n is the first n od no. then (1)-(2); we get the sum of even no.......(3)

thn (3)-(2) we get the ans

2 - 1 = 1; 4 - 3= 1. There are 50 pairs of odd and even numbers from 1 to 100 having the difference of 1. Therefore: 1x50=50

Even numbers: 2+100=102, 4+98=102, etc. There are 25 pairs of even numbers between 1 and 100. 25 x 102=2550. Odd numbers: 1+99=100, 3+97=100, etc. There are also 25 pairs of odd numbers between 1 and 100. 25 x 100=2500. 2550-2500=50

(2 + 4 + 6 + ... + 98 + 100) - (1 + 3 + 5 + ... + 97 + 99) = (2-1) + (4-3) + ... (100-99) = 1 + 1 + 1 + ... + 1 (we have one 50 times). So the answer is 1 x 50 = 50

Each pair of even minus odd = 1
i.e 2-1 = 1 and 4-3 = 1 and 6-5 = 1
Solutions is number of pairs that exist.
Pairs up to 10 = 10/2 or 5, pairs up to 50 equals 50/2 or 25.
Therefore 100 divided by 2 = 50

even=n^2, odd=n(n+1)

From 2 to 100, there are [(100-2)/2+ 1]=50 even numbers. Their sum = [(2+100)/2] x 50= 2550. From 1 to 99, there are [(99-1)/2+1]=50 odd numbers. Their sum = [(1+99)/2] x 50 = 2500. Their difference = 2550-2500 = 50.

From 2 to 100, there are [(100-2)/2+ 1]=50 even numbers. Their sum = [(2+100)/2] x 50= 2550. From 1 to 99, there are [(99-1)/2+1]=50 odd numbers. Their sum = [(1+99)/2] x 50 = 2500. Their difference = 2550-2500 = 50.

As the even positive integers are in A.P(arithmetic progression) , so the sum of A.P will be equal to first term + last term and then multiply by number of terms and then divided by 2 . The same case will be apply to odd positive integers.

• Nós temos duas P.A. (progressão aritmética).

• A primeira possui somente números pares e a segunda somente números ímpares.

• Logo:

-1) 2,4,6...100

-2) 1,3,5...99

• A diferença é dada pela subtração da soma de todos os pares pela soma de todos os ímpares, então aplica-se a fórmula para soma de P.A.

• (a1 + an)n

       2
  

• Para os pares:

• (2 + 100)50 = 2550

        2
  

• Para os ímpares:

• (1 + 99)50 = 2500

      2
  

E subtrai-se os valores obtidos.

• 2550 - 2500 = 50

• E a resposta é 50.

just use Python:

even = []
odd = []
for i in range(1, 101):
    if i % 2 == 0:
        even.append(i)
    else:
        odd.append(i)
print sum(even) - sum(odd)

$(2+4+6+....88+100) - (1+3+5+.....97+99) = (2-1)+(4-3)+(6-5)+....+(100-99) = 50$

sum of all odd numbers =sqr of total odd numbers eg 1+3+5+7+9=5^2=25 in this case number of odd numbers is 50 therefore total sum of odd numbers=50^2=2500 sum of even numbers Formula: N(N+1) How to Find N = (First Even + Last Even)/2 - 1

Example: 2+4+6+ ....... 100 N = (2+100)/2 - 1 = 50 therefore 50*51=2550 difference = 50

2(n(n+1)/2 - [(2(n(n+1)/2)-50]. The first two terms in the expression cancels leading to answer as 50

There are 50 even numbers and 50 odd numbers. Each even number is 1 more than each odd number. Thus, 50.

2-1=1;4-3=1............... and so on when you subtract an odd number from the even number after it you get 1 as the difference there are 100 numbers so 50 ones therefore answer is 50

All positive even integers less than or equal to 100 are : 2,4,6,8,....,100. Then from the formula below, we can find that 100 is the 50th number : Un = a + b (n-1) ,where Un is nth number, a is the first number, and b is the difference between the first number and the second number. Un = a + b (n-1) ==> 100 = 2 + 2 (n-1) . Hence we get n = 50. Whereas the sum of all even positive numbers can be found using the formula below : Sn = n/2 (2a + (n-1) b). Sn is the sum until nth number, n is the nth number, a is the fist number and b is the difference between them. So we get : Sn = 50/2 (2x2 + (50-1)x2) = 2550. Hence, for the odd numbers (1,3,5,7,...99), the 99th is also the 50th but the sum is not the same, Sn=50/2 (2x1 + (50-1)x2) = 2500. So, the difference between the sum must be 2550 - 2500 = 50.

Pair the numbers so that each even number, n, is paired with the odd number, n-1.

n - (n -1) = 1

So the difference in each pair is 1.

There are 50 such pairs.

Therefore the difference is 50.

an(impar) = 1 + (50 - 1)*2 = 99

Sn(impar) = (50(1 + 99))/2 = 2500

an(par) = 0 + (51 - 1)*2 = 100

Sn(par) = (51(0 + 100))/2 = 2550

Sn(impar) - Sn(par) = 2500 - 2550 = 50

2-1=1 and 4-3=1 ........ 100-99=1 so 100/50 times 1 = 50

Total numbers = $100$

So, number of even integers = number of odd integers = $\frac{100}{2}$ = $50$

Sum of positive odd integers = $n^2$ = $50^2$

Sum of positive even integers = $n^2+n$ = $50^2+50$

Difference = $n^2+n-n^2 = n = 50$

we use sequences to solve the problem

for odd numbers sequence will be 1,3,5,7,....,99 sum = ( 1 + 99 ) * 50/2 = 2500

for even numbers sequence will be 2,4,6,8,....,100 sum = ( 2 +100 ) * 50/2 = 2550

the difference will be 2550 - 2500 = 50

2 + 4 + 6 + 8 + … + 100 - (1 + 3 + 5 + 7 + … + 99) = ?

But if you do a bit of rearranging, you get this:

(2 - 1) + (4 - 3) + (6 - 5) + (8 - 7) + … + (100 - 99) = ?

Which of course is really just this:

1 + 1 + 1 + 1 + … + 1 = ?

So the real question is how many (even - odd) terms are there?

There are 50, so the answer is 50.

There are 50 even numbers from 0-100 and 50 odd numbers. If you take an odd number and subtract it from its succeeding even number, it equals 1. Do that 50 times. 1*50 = 50

S1 = 0 +....+ 100 = (0+100)+(2+98)+...+50

S2 = 1 +...+ 99 = (1+99)+...+(49+51) Then S1-S2= 50

Temos 50 pares de 0-100 ( 0,2,4....96 e 98) calculando a soma de todos eles temos:

Sn = (a1 + an)n/2

Sn = (0 + 98) 50/2

Sn = 2450

Temos 50 ímpares de 0-100 (1,3....97 e 99)

Sn = (a1 +an)n/2

Sn = (1+99)50/2

Sn = 2500

assim

2500 - 2450 = 50

Sum of even - Sum of odd = 5000 - 4950

(2,4,6,8,..,100) Sn'= (2 + 100) 50/2 = 2550

(1,3,5,7,...,99) Sn" = (*1 + 99) 50/2 = 2500

Sn' - Sn" = 50

to have the sum of positive even either odd integers less than or equal to $100$ , we use formula : $Sum = ( \frac{number of terms }{2} )$ $( 1^{st} term + last$ $term )$

for this problem, number of terms $= 50$ . If you ask me why, it because between $1$ to $100$ , we have either $50$ odd and $50$ even integers.

$1^{st}$ we find out the sum of positive even integers :

$Even Sum = ( \frac{50}{2} )$ $( 2 + 100 )$

$Even Sum = 25 \times 102$

$Even Sum = 2550$

Next we find out the sum of positive odd integers :

$Odd Sum = ( \frac{50}{2} )$ $( 1 + 99 )$

$Odd Sum = 25 \times 100$

$Odd Sum = 2500$

the last step, we subtract the sum of even and odd positive integers :

$Answer = 2550 - 2500$

$Answer = 50$

So the answer is $50$

1 . How many positive even integers are less than or equal to 100? answer: 50 2 . How many positive odd integers are less than or equal to 100? answer: 50 3 . WHAT"S THE DIFFERENCE BETWEEN THE FIRST POSITIVE EVEN INTEGER AND THE FIRST POSITIVE ODD INTEGER LESS THAN OR EQUAL TO 100?!?!? THIS IS EASY!!!!!! answer: 1 Duh. 4 .(What's the number of positive even integers are less than or equal to 100 \times the difference between the FIRST positive odd integer that's less than or equal to 100 and FIRST positive even integer that's less than or equal to 100) answer: 50 5 .QED the formatting is crazy!

for the first 10 odd integers... 1+3+5+7+9= 25 for the first 10 even integers... 2+4+6+8+10= 30 sum of even- sum of odd = 5 since for every first 10 numbers the difference is 5, so for second set of numbers another 5 and soon.. therefore, 5*10 = 50

{50(2+100)/2}-{1 + 3 + ... + 97 + 99 = [ 100/2 ] ² = 50²} = 2,550 - 2500 = 50

By adding all the even numbers till hundred and then adding all the odd numbers till hundred.After that I Subtracted the sum of even numbers from the sum of odd numbers

sum how much the positive integers

Calculamos 1/10 da Soma do Ímpares , Depois Você Acrescenta Mais 100

1+3+5+7+9 = 25 2+4+6+8+10=30

Assim Dará (30+130+230+330+430+530+630+730+830+930 ) - (25+125+225+325+425+525+625+725+825+925) = 5x10 = 50

2n(n+1) ÷ 2 - [2n(n+1) ÷ 2 - 50]

n = 50

2(50)(50+1) ÷ 2 - [2(50)(50+1) ÷ 2 - 50] = 2550 - [2550 - 50] = 50

Working on a small set of numbers and finding a pattern upto 10 instead of going to 100 can help. Take the numbers in pairs. First pair is (1 and 2), 2-1 = 1 Now take (3 and 4) 4-3 = 1 next (5 and 6) 6-5 =1 , I am seeing a pattern here

7 and 8 --> 1 again

9 ,10 ---> 1 again

We can see that each pair adds one to the sum. With an upper limit of 10, we have 5 pairs. So the sum is 10/2 = 5. Test it by trying the pattern to 20. The pattern works, so the answer is 100/2 = 50

let x1 be sum of even no.s. let x2 be sum of odd no.s.

x1=2+4+6+⋯+100

x2=1+3+5+⋯+99

(2+4+6+...+98+100)−(1+3+5+...+97+99)

2+4+6+...+98+100−1−3−5−...−97−99 We could group them into:

(2−1)+(4−3)+(6−5)+...+(98−97)+(100−99)

=1+1+1+...+1+1 (50 times)

=1×50=50

Ans.=50

100 and 100=50 if divided

2-1 = 1 . . . 100 - 99 = 1 I have 50 times 1

even - odd = 1; 1 x 50 = 50

There are 25 pairs of even integers less than or equal to 100 that add up to 102 so 102 x 25 =2550 Then, there are 25 pairs of odd integers less than or equal to 100 that add up to 100 so 100 x 25 = 2500 therefore, 2550-2500=50

the sum of the odd numbers are 1+3+5+7+...+99 = 50^2 (sine there are 50 terms in the sum and using the formula that 1+3+5+..+n = [(n+1)/2]^2 and the sum of the evens = 2+4+6+8+...+100 = 2(1+2+3+..+50) = 2* 50 * 51 / 2 = 50 * 51 thus the difference = 50 * 51 - 50 * 50= 50(51-1)=50

(50 × 51) - (50^2) =50

I also found that the best way was not to add up all the even numbers and then the odd, but to look at pairs of even and odd numbers, as 2-1=1 (the first applicable pair of numbers), 4-3=1, 6-5=1 etc. As 100/2=50, we will have 50 such sums. 50*1=50, and so the difference is 50.

Of the first 100 positive integers, 50 integers are even and 50 integers are odd. So, sum of first n even numbers=n(n+1) ==> 50*51=2550. sum of first n odd numbers=n^2 ==> 50^2=2500 difference=2550-2500=50.

no. of even integers from 1-100 = 25 no. of odd integers from 1-100= 25 Sum of even integers= 25(2+100)= 2550 Sum of odd integers = 25(1+99)= 2500 2550-2500= 50

the number of even or odd numbers less than 100 is 50

I have solved the problem in $2$ different ways.

• Using $Gauss$ like thinking.

• Using known formulas.

  I am showing them respectively. You may like the first one.

$1)$

Let the sums be $\xi_1$ & $\xi_2$ respectively. Then,

$\;\;\;\;\;\;\;\:\xi_1 = 2+4+6+\dotsb+100$

$\;\;\;\;\;\;\;\:\xi_2 = 1+3+5+\dotsb+\;\;99$

So,

$\xi_1-\xi_2=1+1+1+\dotsb+\;\;\;1=50$

[As there is $50$ outputs as $1$ ]

$Ans.=50$

$2)$

$\xi (considering\;\;sum) = \left\{ \begin{array}{l l} \frac{n(n+1)}{2} & \quad \text{\$n\$ consecutive natural numbers}\\ n^2 & \quad \text{\$n\$ consecutive odd numbers} \end{array} \right.$

Then, we can say, $\xi_1=2\times\frac{n(n+1)}{2}=50\times(50+1)=2550$

[As there are fifty numbers]

And, $\xi_2=n^2=50^2=2500$

Then,

$\xi_1-\xi_2=50$

we know sum of even term =50/2[2 2+(50-1)2]=2550 we know sum of odd term =50/2[2 1+(50-1)2]=2500 so difference=50

For the sum of first n odd number S=(n/2)(2n+1) For the sum of first n even number S=(n/2)(2n+2) for 0<=n so the differnce of the sum of first n odd number and sum of first n even number =(n/2)(2n+2-2n-1) =(n/2)(1) =n/2 therefore for the difference between the sum of all positive even integers less than or equal to 100, and the sum of all positive odd integers less than or equal to 100 =100/2 =50

(2 + 4 + ... + 100) = 50/2(2 + 100) = 2550; (1 + 3 + ... + 99) = 50/2(1+99) = 2500;

2550 - 2500 = 50

I used arithmetic progression method, Sn = n/2 x (2a + (n-1)d)

summation of (n+1 )-summation of (n) = 1, where n=1,3,5,6,7... all give solution = 1 and n<=100 they have 50 such n values so 50 is answer

1 to 100, 50 are even 50 are odd..so by using formula sum = n/2{2a+(n-1)d}, n- total nos, d- difference(in this case -2), a- first no. of series. so sum of even=2550 and sum of odd=2500

Solution, but one that doesn't result as much in a sense of really answering the question: You have to evaluate (2+4+6...100)-(1+3+5....99). Try up to 10, first: 2+4+6+8+10=30 1+3+5+7+9=25, so 30-25=5. 10 10=100, so 5 10=50. The answer is then 50.

The way to solve this for me is following. We divide number from 1 to 100 into pairs (2-1) ( 4-3) (6-5) and so forth. There are 50 pairs. Each pair result is 1. 50*1=50

First we make an arithmetic series: (2,4,6,8,...,100), of the series can be obtained by the equation Un=a+(n-1)b that n=50. same way for series: (1,3,5,7,...,99), and obtained n=50 and then, subtract sum of the series even with an odd sum of the series with the equation Sn=n(2a+(n-1)b)/2 be obtained by the difference is 50

Observe that this is an arithmetic series and there are 50 even/odd integers from 1 to 100. In general,

(a) sum of first N positive even integers is: N(N+1)

(b) sum of first N positive odd integers is: (N)^2

>where N is the number of terms in the series.

Solution: if N=50 then:

(a) = N(N+1) = 50(50+1) = 2550

(b) = (N)^2 = (50)^2 = 2500

Thus, (a) - (b) = 2550 - 2500 = 50.

Answer: 50. ■

The solution is given by the Sum of all odds from 2 to 100 minus the Sum of all even from 1 to 99. We can discover it by doing Sum of 2n from 1 to 50 minus the Sum of 2n - 1(which is always odd) from 1 to 50.

We can group even numbers in 25 pairs that their sum is equal to 100, like (0+100, 2+98, 4+96, 6+94), except for number 50. Also we can group odd numbers in 25 pairs that their sum is equal to 100, like 1+99, 3+97,... Then, if there are 25 pairs that sum 100 in each set but number 50 is alone, then the difference is equal to 50.

We can look at the set of all even numbers less than or equal to 100 as an arithmetic sequence.

The sum of an arithmetic sequence is $S = \frac{n}{2}(t_1 + t_n)$ , where $n$ is the number of terms, $t_1$ is the first term, and $t_n$ is the last term. There are 50 even numbers from 1 to 100, the first one being 2 and the last one being 100. Using the formula, we conclude that the sum of all even integers less than or equal to 100 is 2550.

We can interpret the set of all odd integers less than or equal to 100 in the same way. Using the same formula, with $n = 50$ because there are 50 odd integers in the sequence, $t_1 = 1$ , and $t_n = 99$ , we get that the sum of all odd numbers less than or equal to 100 is 2500.

2550 - 2500 = 50.

First of all, we need to take the numbers in a tabular form:- 1 + 3 + 5 + 7 + 9 + 11 +......... 2 + 4 + 6 + 8 + 10 +12 +... By the reference of even numbers, at each set of 5, the net difference is 5. So, since there are 10 more such sets of 5, we get the difference * 10 = 50. I took 50 because odd + even =100. So, sum there should be total 50 even numbers in the series of 1 to 100.

As the even number is the final number, it is easiest to count down: (100+98+96...+2) - (99+97+95...+1) = (100-99) + (98-97)...+(2-1) = 50(1) = 50

1,3,5..................99(50 terms) by arithmetic progression, sum of n terms, s=n/2(2a+(n-1)d), similarly 2,4,5.................100

Sum of odd=1+3+5...+97+99 Sum of even=2+4+6...+98+100 Difference =(100-99) + (98-97) +... (4-3) + (2-1) = 50(1) = 50

Ans=(2+4+6...+100)-(1+3+5...+99) [50 terms in each] =(2-1)+4-3)+...+(100-99) =1+1+1...+1 =50

see 2-1=1;(2+4)-(1+3)=2;(2+4+6)-(1+3+5)=3......upto 100 since upto 100 the total number of odd and even is 50,50.so the answer can easily get

difference between odd and even is always 1. There are equally number of odd and even in 100. therefore 50*1 = 50

Take the highest even number less or equal to 100 and the lowest even number higher than 0.(100+2=102;98+4=102........) since you are dividing a hundred numbers into pairs,divide hundred by two. Now, since you are using even numbers,divide fifty by two. You get twenty five. Multiply 102*25=2550.

Now do the same thing with odd numbers.(1+99; 3+97....) you get 25 pairs of 100's. 100*25=2500.

Subtract 2000 from 2550.

2-1=1, 4-3=1........................

So, it would repeat 50 times as there are 50 pairs.

So, 50X1

=50

As we know that, 2, 4, 6, ... and 1, 3, 5, ... are arithmetic series. Sum of arithmetic series is equal to Sn = n/2 (1st term + nth term) where n is the number of terms. So, 2 + 4 + 6 + ... + 100 = 50/2 (2 + 100) = 2550 And, 1 + 3 + 5 + ... + 99 = 50/2 (1 + 99) = 2500 So the difference will be, 2550 - 2500 = 50 Hence 50 is the required answer.

as positive even numbers less than or equal 100 form a seqeunce of < 2 ,4,6 ,....100> of T1 =2 number=50 difference =2 and last term =100 ,, so sum equal n/2 (2*T1 +(n-1)d) which equals 50/2 * (4 + 98 ) =2550 similarly the seqence of positive odd numbers ( 1, 3 5 , ....99 ) which has a sum of 2500 so difference equal 2550 - 2500 = 50

All positive even integers are = 2,4,6,......100. It is in a form of a.p. where difference b/w each consecutive term is constant. So by the formula of ap i.e. An=A1+ (n-1) d, where An=100, A1=2, d(common difference)=2 so n(Total no. of terms)=50 Hence the sum of an ap is represented as Sn= n/2[2A+(n-1)d] = 50/2 [2 2 + (50-1) 2] so the sum of all positive even integers is 2550. Similarly all positive odd integers= 1,3,5,.....99 it is also in form of ap with a common difference of 2. so we calculate the no. of terms by same formula An=A1+ (n-1) d , where An=99, A1=1, d(common difference)=2 so n(total no. of terms)=50 And the sum of this a.p. by Sn= n/2[2A+(n-1)d] = 50/2 [2 1 + (50-1) 2] hence the sum of all positive odd integers is 2500. So the difference between the sum of all positive even integers less than or equal to 100, and the sum of all positive odd integers less than or equal to 100 is = 2550-2500 = 50

(2+4+6+...98+100)-(1+3+5+...+97+99)=(2+100).50/2-(1+99).50/2=2550-2500=50

So if you take the sums of the first 10 even and odds (2+4+6+8+10) and (1+3+5+7+9) you will get a difference of 5, which is also the case if you do it from 11-20. If there are 10 sets of 10 in 100, then you will have 5 times the 10 sets giving you the 50 since the pattern holds true throughout.

This way is a bit long, but it works... There are 50 positive even integers ≤ 100, and 50 positive odd integers ≤ 100.

For the even integers, add the first term (2) to the last term (100) to get 102. Then multiply the number of terms in the series (50) by 102 to get 5100. Divide 5100 by 2 to get 2550, which is the sum of the even integers ≤ 100.

Do the same thing with the odd integers. So, 1+99 = 100, and 50*100 = 5000. 5000/2 = 2500, which is the sum of the odd integers ≤ 100.

Finally, subtract 2500 from 2550 to get the difference between the sums, which is 50.