Alternating back and forth

(100-99)+(98-97)+............(2-1) 1+1+1+.......+1 = 50

We can to separate the equation in groups (100-99) + (98-97) + ... + (4-3) + (2-1) We have 50 groups. The first has the number 2, the second has the number 4 ... the 50° has the number 100. The sum in ever brackets is 1. The sum of the results of all brackets is 50.

Sum of n even numbers =n(n+1).... Sum of n odd numbers = nxn... sum of n even numbers - sum n odd numbers = n(n+1)-nxn=n.... Here we have n value is 50.. .. So answer is n=50.

100 - 99 + 98 - 97 + 96 - 95 ....... + 4 - 3 + 2 - 1 = (100 - 99) + (98 - 97) + (96 - 95) + ..... + (4 - 3) + (2 - 1) here we can see that its a 50 numbers of 1 added to each other then the value is 50

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               100, 98, 96, ...... 6, 4, 2

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               - 99, 97, 95, ...... 5, 3, 1

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                = 1,  1,  1, ...... 1, 1, 1

Number of terms of the sequence above, n is 50. Hence, sum of the sequence, S is

S = n(1)

S = 50(1)

S = 50

100/2= 50 Karena menurut logika saya setiap 2 hasil (contoh 100-99=1 dan 1 + 98 = 99 dan seterusnya ) akan menghasilkan 100. hehe in my opinion. jadi 100 dibagi 2 saja. biar cepat.

We know that 2-1 = 1 ; so if we repeat this sequence to 100-99 = 1 we will be repeating the sequence for 50 times , and 50*1=1 . then we conclude that the sum is 50.

=(100-99)+(98-97)+(96-95)+(94-93).......................+(4-3)+(2-1) [total 50 pairs] =1+1+1+1+1+............+1+1 =50

The question is 100 - 99 + 98 - 97 ... +2 - 1

First consider the first two part 100 - 99 = 1

Then, second part 98 -97 = 1

Similarly, if you notice all answers for all two parts is 1

Since, there are 100 numbers altogether and we are taking two numbers at a time

So, there are 50 answers means 50 1s.

Adding all we will get 50 as final answer.

I think this is the fastest and most easy way to solve these problems

I hope this will help you.

Thank You !

(100-99)+(98-97)+(96-95)+...+(2-1) = (1)+(1)+(1)+...+(1) = 50.

The terms consists of summation of negative odd numbers and positive even numbers Therefore their nth term is (1-2n) for the negative terms and (2n) for the positive terms. The sum of these terms is (1-2n+2n) which is =1. Since there are 50 terms each their sum becomes 50 x 1=50

100 - 99 + 98 - 97 + 96 - 95 ....... + 4 - 3 + 2 - 1 = (100 - 99) + (98 - 97) + (96 - 95) + ..... + (4 - 3) + (2 - 1) here we can see that its a 50 numbers of 1 added to each other then the value is 50

Cara gaul and asik abiz -> 100 - (100-1) - 98 - (98-1) - ... - 2 - (2-1) :ngakaks ; kaskus = rafi.rp

Let S = 1+3+5+...+97+99, Then S + 50 = 2+4+6+..+98+100. (Adding 1 to each number, and there are 50 numbers in total). Therefore(S+50) - S = 50 is the answer

100 - 99 + 98 - 97 + 96- 95 +.....+ 4 - 3 + 2 - 1

=(100 - 99) + (98 - 97) + (96 - 95) + .......+ (4 - 3) + (2 - 1)

each of these brackets is equal to 1 , and the whole sequence is divided into pairs , then:

sum= 1x (100/2 ) = 50

=(100-99)+(98-97).........+(4-3)+(2-1) =1+1+1.....+1+1(total 100/2) = 50

(100 - 99) + (98 - 97) + .... = 1 + 1 + ..... until 50 times. so, 1 x 50 = 50

as =(100 - 99)+(98 -97)+(96 - 95)+,,,,,,,,,,,+(2 - 1) =1+1+1+1,,,,,+upto 50 times =50

100 - 99 =1 98 - 97 = 1 96 - 95 = 1 continuing this process and from 1 to 100 there are 50 pairs so the sum of each pair is 1. adding them all is 50.

sum of n even no. is n.(n+1) and sum of n odd no. is n^2. i.e. here even no. are 50 and odd no. are 50. hence n.(n+1)-n^2= 50.(50+1)-50^2=50

Combining the greatest and the least negatives make the additive inverse of the even number on the left side of the greater minuend.

-99 + -1 = -100

-100 + 100 = 0

Same goes with

-97 + -3 = 100

98 + 2 = 100

-100 + 100 = 0

But at some point in the middle leaves one number retained. The middle number is 100/2 = 50

52 + 48 = 100

-51 + -49 = -100

50.

It is to see that every pair is 1,and you have 50 pairs, so 50X1=50

As we observe we have 100 integers......... every doublet(group of 2) is having difference of two........... total doublets=100/2=50............... therefore answer=50x1=50

sum of first 50 odd numbers: odd sum= (50/2)((2*1)+ (50-1)(2))= 2500 sum of first 50 even numbers: even sum= (50/2)((2*2)+ (50-1)(2))= 2550 thus, as per question, value= 2550-2500=50

There are two arithmetic progressions in the given sum of series.

First is: 100, 98, 96, 94, ………, 2

Second is: 99, 97, 95,…….., 1

We first find the number of terms in each series by the formula:

an = a1 + (n-1)d

where an is the last term and a1 is the first term of series and d is the common difference in the terms.

First series: a1=100, an=2 and d=2

Therefore, 2= 100 + (n-1)2

Therefore, 2-100 = (n-1)2

Therefore, -98 = (n-1)2

Therefore, (-98)/2 = n-1

Therefore -49 = n-1

Therefore Neglecting negative sign,

Therefore 49 = n-1

Therefore n=49+1

Therefore n=50

In the same manner, n=50 for second series also, where an=1, a1=99 and d=2

Sum of first series=n(a1+an)/2

=50(100+2)/2

=25(102)

=2550

Sum of second series=n(a1+an)/2

=50(99+1)/2

=25(100)

=2500

Therefore, 100-99+98-97+96-95+……….+4-3+2-1

Therefore, (100+98+96+……+4+2)-(99+97+95+……+3+1)

=(2550-2500)

=50

between 1and 10 :

10-9 =1 8-7= 1 6-5 = 1 4-3 =1 2-1 =1 Total sum : 5

Between 1-100 there are :

5*10 =50

the additive and subtractive terms form an AP. So By using the formula N/2(A+L) where, n=no of terms, a=first term, l=last term the answer comes down to: (50/2) (2+100) - (50/2)(1+99)=50

(100-99)+(98-97)+(96-95)+.........+(4-3)+(2-1)=1+1+1.....+1+1=50{the easy way to solve it without using A.P.!!}#commonsense

100-99+98-97+96.....+2-1

= (100+98+96+....+2) - (99+97+95+...+1)

= ((99+1)+(97+1)+(95+1)+....+(1+1)) - (99+97+95+...+1)

= (99+97+95+....+1) + (1+1+1+...upto 50 terms) - (99+97+95+...+1)

= (1+1+1+...upto 50 terms) = 50

I noticed there was a pattern: every single pair of subtraction is equal to 1. We have 50 pairs, because another pattern: the count never makes reference to the same number twice, then, if the count starts at number 100 until 1, we will have 50 pairs. To calculate the answer is pretty easy if you notice this to patterns: 50 x 1 = 50. Use the same principle that we use to calculate the sum of a A.P.

100-99=1, 98-97=1 ,
96-95=1, 94-93=1, 92-91=1, hence 1-10=5, 1-100=5*10=50,

1+1+1+...+1 --> sebnyak 50 = 50