n=2

The sequence for 2nd term is 2²+1 = 5
The sequence for 3rd term is 3²+1 = 10
The sequence for 4th term is 4²+1 = 17 etc

The value of the (n+1)th term (3rd term) minus the value of the nth term (2nd term)
10 - 5 = 5 which is equal to (2 x 2) + 1

sofiasol wrote:A sequence generated by the rule that the kth term is k²+1 for each positive integer k. In this sequence, for any value of n>1, the value of the (n+1)th term less the value of the nth term is


a) 1
b) 3
c) n²
d) 2n+1
e) n²+1

Maybe the best thing to notice first is that since each term has 1 added to it, in calculating the differences, we can ignore the 1's.

In other words, (y² + 1) - (x² + 1) = y² - x²

So basically what we are working with are the differences between the squares of successive integers.

One way to get to the answer, therefore, is to just plug some consecutive integers into the sequence and then see which of the answer choices we can eliminate.

2² = 4
3² = 9
4² = 16
5² = 25

Since the differences between the squares are all greater than 3, we can eliminate A and B.

Since 2² + 2² ≠ 3², we can eliminate C.

Let's try D.

2² + 2(2) + 1 = 9 = 3²

3² + 2(3) + 1 = 16 = 4²

D is looking good so far. So let's eliminate E, if possible.

3² + 3² + 1 = 19 ≠ 4²

So E is out.

The correct answer is D.

Now, there is a much faster way to do this. Vision rules on the GMAT, and so by seeing something, you can burn through this question.

We are looking for the difference between n² + 1 and (n + 1)² + 1 which is the same as

(n + 1)² - n².

You can multiply out the first term to get the following.

(n² + 2n + 1) - n² = 2n + 1

The correct answer is D.

Hi Rich

I agree. I will be careful next time.

(sofiasol: Sorry for the inconvenience. The approach is correct so you can still use it but just remember to test with different numbers if you have two options yielding same answer).

Thanks

Let's make it a function instead, such that

f(k) = k² + 1

We're asked to find f(n + 1) - f(n), so we'd get

f(n + 1) = (n + 1)² + 1

and

f(n) = n² + 1

Then simply do (n + 1)² + 1 - (n² + 1).