Hello it’s my first time posting to please bear with me. This question #8 on page 47 in the ETS quant book in the arithmetic section.

For each integer n>1, let A(n) denote the sum of the integers from 1 to n. For example, A(100)=1+2+3…+100=5050. What is the value of A(200)

A. 10,100

B. 15,050

C. 15,150

D. 20,100

E. 21,500

My first instinct was to double the 5050 but that is incorrect. I don’t really understand the ETS explanation of choice D. Any help is greatly appreciated!

It’s good to have some formulas handy. Sum of the first n natural numbers can be written as \Sigma n = \frac{n(n+1)}{2}

Even if you don’t know the formula, if you know that A(100) = 5050, just think about it like this.

- 1+2+3…100 = 5050
- 1+2+3…200 = (1+2+3…100) + (101+102+103…200)

So if you see the second equation there are two summations, one from 1 to 100 and one from 101 to 200. So doubling the answer will not give you the answer because there is an extra 100 in each of the term of the second sequence. So there are hundred 100s extra (which add up to 10000).

- 1+2+3…200 = (1+2+3…100) + (100 + 1 + 100 + 2 + 100 + 3… 100+100)
- 1+2+3…200 = (1+2+3…100) + (100 + 100+100… 100 times) + (1 + 2 + 3…100)

From 101 to 200 we are doing 100+1,100+2,100+3,100+4…100+100

If it was doubled as you have said then we would have 1+2+3+4 and not 100+1(101),100+2(102),100+3(103),100+4 (104)etc…

So, if you wanna do it by your logic just account for that extra hundreds from 101 to 200, which are 100*100 + (5050)*2[because if we remove the hundreds then we are left with 1+2+…+100] = 20,100

Thank you so much! This makes sense I didn’t think about the extra 100 in 200.

Thank you so much! I will be committing this equation to memory.