Answer:
Step-by-step explanation:There are formulas to work this out, but you can also suss out an answer by a liberal application of critical thinking.
Okay, you want to add 51 + 52 + 53 + … + 99 + 100, right? (The … represents all the numbers in between, in case you were wondering.)
Well, that’s a total of 50 numbers. And if you know anything about addition, you know that it doesn’t matter what order you add them.
So you could also say 51 + 100 + 52 + 99 + 53 + 98 + … and so on.
Let’s see what happens. 51 + 100 = 151. So far so good.
Then, 52 + 99 also equals 151. And 53 + 98 also equals 151. Since there are 50 numbers, that’s 25 pairs that are going to add up to 151.
151 + 151 + 151 + 151 + … (21 more instances of 151).
So we can simplify our workload a bit. Instead of adding up all the numbers by hand, we can just multiply 151 by 25.
151 x 25 = 3775
From that, we can work out a generic formula that will allow you to add any sequence of numbers…without actually adding.
Let a be the first number in the sequence, and let n be the last number in the sequence. The number of numbers is (n - a) + 1. (You have to add 1, because if you just subtract, it’s like you’re losing the lowest number in the sequence).
So…you get the sum by adding the first and last number in the sequence, and then multiplying that result by half of the total number of numbers.
(a+n)((n−a)+12)
an−a2+a+n2−an+n2
−a2+a+n2+n2
−a(a−1)+n(n+1)2
Let’s check to see if it works!
−51(51–1)+100(100+1)2
−51(50)+100(101)2
−2550+10,1002
75502
3775
Note that if the starting number were 1, the formula would simplify to n(n+1)2
![2[cos(A-B)+1]](https://tex.z-dn.net/?f=2%5Bcos%28A-B%29%2B1%5D)