Q: What is the sum of 1 plus 2 plus 3. plus 100?

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The sum of the numbers up to 100 is equal to (100 x 101)/2 = 5050

The mater is that if you do 1+1=2 then just add the sum with its self, so you have 2+2=4 and 4 is the answer

100 + 99 + 98 + 2 + 1 + 0.3 = 300.3

Consider a denominator of r; It has proper fractions: 1/r, 2/r, ...., (r-1)/r Their sum is: (1 + 2 + ... + (r-1))/r The numerator of this sum is 1 + 2 + ... + (r-1) Which is an Arithmetic Progression (AP) with r-1 terms, and sum: sum = number_of_term(first + last)/2 = (r-1)(1 + r-1)/2 = (r-1)r/2 So the sum of the proper fractions with a denominator or r is: sum{r} = ((r-1)r/2)/r = ((r-1)r/2r = (r-1)/2 Now consider the sum of the proper fractions with a denominator r+1: sum{r+1} = (((r+1)-1)/2 = ((r-1)+1)/2 = (r-1)/2 + 1/2 = sum{r) + 1/2 So the sums of the proper fractions of the denominators forms an AP with a common difference of 1/2 The first denominator possible is r = 2 with sum (2-1)/2 = ½; The last denominator required is r = 100 with sum (100-1)/2 = 99/2 = 49½; And there are 100 - 2 + 1 = 99 terms to sum So the required sum is: sum = ½ + 1 + 1½ + ... + 49½ = 99(½ + 49½)/2 = 99 × 50/2 = 2475

Sum of first n natural numbers is (n) x (n + 1)/2 Here we have the sum = 100 x (101)/2 = 50 x 101 = 5050

Related questions

5050

303

The sum of the numbers up to 100 is equal to (100 x 101)/2 = 5050

Sum of numbers from 1 to 100:(1+100) + (2+99) + (3+98) + (4+97) + ... + (49+52) + (50+51) =50 pairs, each pair adds up to 101= (50 x 101)= 5,050

2*100 = 200 12*10 = 120 14*1 = 14 Sum = 334

You add the numbers in a loop. Here is an example in Java:int sum = 0;for (int i = 1; i

if (n%2==0) sum=n/2*(n+1); else sum=(n+1)/2*n;

Sum = 79*(79+1)/2 = 79*80/2 = 3160

The sum of these composite numbers, including 100, is 3989.The sum of all numbers between 1 and 100 is 5,050. From this we need to subtract the sum of 1 plus all the prime numbers below 100. The sum of the primes is 1,060. Subtracting (1 + 1060) or 1,061 from 5,050 yields 3,989.

2

2

1+2+4 = 7