Give a rule for finding the first of four consecutive odd integers if you know the sum of the four integers.
2 answers:
This ones a bit of a trick question, since whether the integers are even or odd doesn't really matter. What does matter, is that consecutive odd integers are 2 numbers apart (as are consecutive even integers).
So again, if the first of the odd integers is the variable n
the consecutive odd integers after it can be written as:
n+2 , n+4, etc...
So the sum of four consecutive odd integers can be written as:
sum = n + (n+2) + (n+4) + (n+6)
Simplify:
sum = 4n + 12
And finally rearrange to solve for n:
= n
This can also be written as:
- 3 = n
Whichever way you prefer.
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The easy thing to do is notice that 1^4 = 1, 2^4 = 16, 3^4 = 81, and so on, so the sequence follows the rule . The next number would then be fourth power of 7 plus 1, or 2402.
And the harder way: Denote the <em>n</em>-th term in this sequence by , and denote the given sequence by
.
Let denote the <em>n</em>-th term in the sequence of forward differences of
, defined by
for <em>n</em> ≥ 1. That is, is the sequence with
and so on.
Next, let denote the <em>n</em>-th term of the differences of
, i.e. for <em>n</em> ≥ 1,
so that
etc.
Again: let denote the <em>n</em>-th difference of
:
etc.
One more time: let denote the <em>n</em>-th difference of
:
etc.
The fact that these last differences are constant is a good sign that for all <em>n</em> ≥ 1. Assuming this, we would see that
is an arithmetic sequence given recursively by
and we can easily find the explicit rule:
and so on, up to
Use the same strategy to find a closed form for , then for
, and finally
.
and so on, up to
Recall the formula for the sum of consecutive integers:
and so on, up to
Recall the formula for the sum of squares of consecutive integers: