Sequence question

[zofaan]

Here is a question I wrote. I want to make sure it's correct.

The first terms of an infinite sequence are 1, 2, 4, 7, 11, ... and continue such that the differences between subsequent terms continue to increase by 1. What is the 100th term in this sequence?
(A) 4851
(B) 4950
(C) 4951
(D) 5050
(E) 5051

 

[Ishuda]

Well, a simple explanation is that is that the difference between the nth and (n-1)st term is n, i.e.
a_n= a_n-1 + n
= a_n-1 + n
= a_n-2 + (n-1) + n
...
= a₁ + ?
Work it out or highlight between the >>>... and ...<<<

>>>>>>>>>>>>>>>>>>>>>>>>>
= a_n-3 + (n-2) + (m-1) + n
= ...
= a₁ + (n - (n-1)) + (n-(n-2)) + ... + n
= a₁ + 1 + 2 + 3 + ... + n
= a₁ + n (n-1) / 2
<<<<<<<<<<<<<<<<<<<<<<<<<
Thus
a₁₀₀ = a₁ + ?
= ?

 

[zofaan]

I'm looking for the more elegant solution.

 

[HallsofIvy]

In other words, you didn't understand what Ishuda meant. You are given that the differences increase by 1- that is, the "second difference", the "differences of the differences" is 1. It follows that the formula is quadratic.

Assuming that a_n= A+ Bn+ Cn^2, a_1= A+ B+ C= 1. a_2= A+ 2B+ 4C= 2, and a_3= A+ 3B+ 9C= 4.

Solve those three equations for A, B, and C. Then check that you get the given values for n= 4 and 5 and finally use that to calculate a_100.

 

[Ishuda]

I'm looking for the more elegant solution.

Well why didn't you say so at first. The 100th term is the fifteenth word on page 12 in the fifth part of the First Edition of 'A Compendious System of Anatomy. In Six Parts'. Or, if that is still not elegant enough for you, I think I will have to let you do your own work.