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1+2+3+4.....+100 =

101*50= 5050

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โˆ™ 2011-02-15 02:47:13
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A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

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

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


What is 1 plus 1 plus 1 plus 1?

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


What is 100 plus 99 plus 98 plus 2 plus 1 plus 0.3?

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


What is the sum of the numbers from 1 to 100?

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


What is the sum of all the positive proper fractions with denominators less than or equal to 100?

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

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