1
$\begingroup$

In terms of $n$, find the formula for the sequence $a_n$: 7, 10, 16, 25; given that the difference between each adjacent element in the sequence creates an arithmetic sequence of its own.

I got the incorrect solution and I cannot trace my mistake:

We create a new sequence $b_n$ based on the given information:

$b_1 = 10 - 7 = 3$
$b_2 = 16 - 10 = 6$
$b_3 = 25-16 = 9$
$d_{b_n} = 3$

We know that: $$a_n - a_{n-1} = b_{n-1}$$ If we were to add all possible elements: $$A_n - A_{n-1} = B_{n-1}$$ We also know that $A_n - A_{n-1} = a_n$, so therefore $a_n = B_{n-1}$: $$a_n = \frac{[2b_1 + d(n-2)](n-1)}{2} = \boxed{\frac{3n(n-1)}{2}}$$ However, this is incorrect as it gives values inconsistent with the definition of the sequence. The correct answer is $\frac{3n^2-3n+14}{2}$.

Where have I gone wrong?

$\endgroup$
  • $\begingroup$ I don't see where you used the initial condition $a_1=7$. Just knowing the differences between terms in a sequence does not determine the sequence uniquely. You still need some more information, like a starting value. $\endgroup$ – lulu Jan 11 '19 at 17:05
  • $\begingroup$ @lulu Isn't it implied from $b_1 = 10 - 7 = 3$? $\endgroup$ – daedsidog Jan 11 '19 at 17:07
  • $\begingroup$ No. Just knowing that $a_2-a_1=3$ doesn't tell you what $a_1,a_2$ are. $\endgroup$ – lulu Jan 11 '19 at 17:17
2
$\begingroup$

You have an error when adding the sequences up to $n$.

You have the following:

$$ a_2 - a_1 = b_1 \\ a_3 - a_2 = b_2 \\ \dots \\ a_n - a_{n - 1} = b_{n - 1} $$

So, when you add these equalities, you get:

$$\sum_2^n a_i + \sum_1^{n-1}a_i = \sum_1^{n-1}b_i$$

The second term here is indeed $A_{n-1}$, and the right hand side is $B_{n=1}$, but the first is $A_n - a_1$. Therefore you have $A_n - a_1 - A_{n-1} = B_{n-1}$. Now, using $A_n - A_{n-1} = a_n$, we have $$a_n = B_{n-1} + a_1 = \frac{3n(n-1)}{2} + 7$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.