What are the numbers that we add together called?
DIFFERENT TERMS INVOLVED IN ARITHMETIC OPERATIONS.If you add a and b, they are called the ADDENDS. The result of a+b is called the SUM.If you subtract b from a, a is called the MINUEND and b is called the SUBTRAHEND. The result is called the DIFFERENCE.If you multiply a and b, they are called MULTIPLICANDS. The result of a*b is called the PRODUCT.If you divide a by b, a is called the DIVIDEND and b is called the DIVISOR. The results of division are the QUOTIENT and the REMAINDER.If a number a is multiplied by itself b times, then a is called the BASE and b is called the EXPONENT or POWER.If you take nth root of a number a, then a is called the RADICAND and n is called the INDEX. √ sign is called the RADICAL.
Add All These Numbers together ?
i like to add the easy numbers first (since it doesn't matter what order you add them in). so i add 300 + 200 + 1000 = 1500, so far. then the not-so easy numbers together: 250 + 450 = 700. add this to the 1500, to get 2200, so far. adding 470 isn't so bad, either: 2200 + 470 = 2670, so far. just one number left to go, adding 526. ok, deep breath: 2670 + 526 = 2670 + 500 + 26 (sneaky, huh?) = 3170 + 26 = 3196. (since every step was easy, there's not as much chance that i made a mistake).
If you add two prime numbers together, will the sum also be a prime number?
All prime numbers, with the exception of 2, are odd. If you add two odd numbers together you get an even number; none of which can be prime because they will be divisible by 2. However, if one of the primes is 2 then the only time adding two primes together would result in another prime is if the prime number was a part of a twin prime. A twin prime are two primes number whose difference is 2. eg. 3, 5 11, 13 17, 19 29, 31 Obviously, adding 2 to the former of a twin prime will result in the latter. Multiplying two prime numbers together will never result in a prime number because the resulting number will have the two prime numbers as factors. If A and B are both prime numbers, A x B = C then C will be divisible by both A and B.
If you add pi together, will it be the biggest number?
If you add pi together, will it be the biggest number?If I take your question to mean:If I add all of the digits in the decimal expansion of [math]\pi[/math] will it be the largest number?(your formulation is unclear), then the answer will be no bigger and no smaller than the result of:[math]1+1+1+1+1+\ldots[/math]Both sums diverge to infinity. Though the second sum may seem less mystical to you.
If I could add all of the natural numbers together, do I actually get -1/12 or is this not true? If true, how is it achieved?
Here is a simple example of how it is achieved.Take the infinite series defined by [math]\,T_k=x^k\,[/math] and the series of partial sums[math]\quad\displaystyle S_n\ =\ \sum_{k=0}^{n}T_k\ =\ \sum_{k=0}^{n}x^k[/math]Each term [math]S_n[/math] is the sum of a simple geometric progression and we know the formula for this is[math]\quad\displaystyle S_n\ =\ \frac{1-x^{n+1}}{1-x}[/math]If [math]\,|x|<1\,[/math] we can find the limit of [math]S_n[/math] as [math]\,n\to\infty\,[/math]:[math]\quad\displaystyle \lim_{n\to\infty}S_n\ =\ \frac{1}{1-x}[/math]and this is because the [math]\,x^{n+1}\,[/math] term tends to zero.This gives me a lovely equality which is valid for [math]\,|x|<1\,[/math]:[math]\quad\displaystyle \sum_{k=0}^{\infty}x^k\ =\ \frac{1}{1-x}[/math]I have a simple expression which agrees with my infinite sum for those values of [math]x[/math] for which that sum converges.I now plug in [math]\,x=2.\ [/math] [Why should I not do this?][math]\quad\displaystyle \sum_{k=0}^{\infty}2^n\ =\ \frac{1}{1-2}[/math]In other words:[math]\quad 1+2+4+8+16+32+\cdots\ =\ -1[/math]This is not dissimilar to the result you quote in the question and the method of achieving this result is identical.The sum of the Dirichlet series is given by[math]\quad\displaystyle D(s)\ =\ \sum_{n=1}^{\infty}\frac{1}{n^s}[/math]and this infinite sum converges for complex [math]s[/math] whenever the real part of [math]s[/math] is greater than [math]1[/math].The Riemann zeta function [math]\,\zeta(s)\,[/math] agrees with the infinite sum [math]\,D(s)\,[/math] for those values of [math]s[/math] for which that sum converges. In fact, it was designed to do so.I now plug in [math]\,s=-1.\ [/math] [Why should I not do this?][math]\quad\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^{-1}}\ =\ \zeta(-1)[/math]In other words:[math]\quad 1+2+3+4+5+6+\cdots\ =\ -\frac{1}{12}[/math]Same trick. Same flaw in logic: The expression only matches the infinite sum when the series of partial sums converges to a limit.
What is every number from 1 to 500 added together equal?
To sum the numbers: 1 + 2 + 3 + ... + n, the formula is n(n+1)/2 In your case, n = 500 500(500+1)/2 = 500(501)/ 2 = 125,250
How Do I Add Sequences Together?
You did what Euler did as a little kid when he added from 1 all the way to 100 in 2 minutes. You pair the first with the last, the second with the second last, and so on until you've paired everything. You find the sum of one pair and multiply it by the number of pairs to find the total sum. 10 + 110 + 210 + 310 = (10 + 310) + (110 + 210) ...................................= 320 + 320 ...................................= 640 20 + 30 + 40 + 50 = (20 + 50) + (30 + 40) .............................= 70 + 70 .............................= 140 640 + 140 = 780 1 + 3 + 5 + 7 2 pairs (4 numbers) 1 + 7 = 8 8 * 2 = 16 1 + 3 + 5 + 7 = 16 20 + 30 + 40 + 50 2 pairs (4 numbers) 20 + 50 = 70 70 * 2 = 140 20 + 30 + 40 + 50 = 140 Keep going.
Add up all of the numbers between 1 and 100...?
Karl Frederick Gauss x= 1 + 2 +..+ 99 + 100 x = 100 + 99 +..+ 2 + 1 so 2x = 101 + 101 +...+ 101 (100 times) = 10100 hence x = 10100/2 = 5050 (Works for any integer n)
What is the sum of all the numbers from 1-500?
Lets start with a small one: The numbers from 1 to 10 add up to 1+2+3+4+5+6+7+8+9+10=55 But you could 'pair' the 1 with 10, the 2 with 9, etc, and create 5 pairs that each add up to 11 11*5 = 55 So, for 1-500, you have 250 pairs (1+500, 2+499, etc) that each add up to 501, for 250*501 in total (or 125250). Generally, for all of the numbers from 1 to n (n being even), the sum will be = n/2 * (n+1) For the ODD numbers, lets go back to 10 again: 1+3+5+7+9=25, but by pairing, we would have '2 and a half' pairs that add up to 10, so that 2.5*10=25 In general, for odd numbers: Sum= n/2*(n)/2 = n^2 / 4 For 500, its 500^2/4 = 62500
What is the sum of all numbers from 1 to 1,000,000?
Easiest way to find the sum of , 1 to 100, 1 to 1000…….1 to 1,000,000 and so on…………For Sum of 1 to 10divide 10 by 2 you will get 5, now write this 5 two times = 55For Sum 1 to 100divide 100 by 2 you will get 50, now write this 50 two times = 5050..For Sum 1 to 1,000,000divide 1,000,000 by 2 you will get 500000, now write this 500000 two times =500000500000