Summats Clear
Find the sum, f(n), of the first n terms of the sequence: 0, 1, 1,
2, 2, 3, 3........p, p, p +1, p + 1,..... Prove that f(a + b) - f(a
- b) = ab.
Find the sum, $f(n)$, of the first $n$ terms of the sequence: \begin{equation*} 0, 1, 1, 2, 2, 3, 3, \dots , p, p, p +1, p + 1, \dots \end{equation*}
Go on to prove that $f(a + b) - f(a - b) = ab$, where $a$ and $b$ are positive integers and $a > b$.
Investigate special cases for small $n$ first.
Can you spot patterns in the sums for $n=3$, $4$, $5$, $6$, $7 \dots$?
Vassil from Lawnswood School, Leeds, Michael from Madras College St Andrews and Koopa Koo from Boston College all solved this problem, well done all of you.
Here is Vassil's solution:
Let $f(n)$ denote the sum of the first $n$ terms of the sequence $$0, 1, 1, 2, 2, 3, 3,\ldots , p, p, p+1, p+1,\ldots.$$
First I tried with several numbers. Let $n=15$. Then $f(15)=2 \times (1+2+3+4+5+6+7)=7 \times 8$
where the sequence is: 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7.
Let $n=14$. Then $f(14)=2 \times (1+2+3+4+5+6+7)-7=7 \times 7$
where the sequence is: $0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7$.
Let $n=17$. Then $f(17)=2 \times (1+2+3+4+5+6+7+8)=8 \times 9$
where the sequence is: $0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8$.
Let $n=16$. Then $f(16)=2 \times (1+2+3+4+5+6+7+8)-8=8 \times 8$
where the sequence is: $0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8$.
I noticed that the formula for $f(n)$ depends on whether $n$ is odd or even.
${\bf Case}$ ${\bf I}$ - $n$ is odd, i.e. $n=2k+1$
Then $$\eqalign { f(n) = 2(1+2+...+k)\cr = 2k(k+1)/2 \cr = {\left(n - 1\over 2\right)}{\left(n+1\over 2\right)}.}$$
${\bf Case}$ ${\bf II}$ - $n$ is even, i.e. $n=2k$
$$\eqalign { f(n) = 2(1+2+...+k) - k \cr = k^2 + k - k \cr = k^2 \cr = \left({n\over 2}\right)^2.}$$
Now we have to calculate $f(a+b)-f(a-b)$.
There are two cases. In the first case, when one of $a$ and $b$ is even and the other is odd, then $(a+b)$ and $(a-b)$ are both odd. Otherwise $(a+b)$ and $(a-b)$ are both even.
Case I $(a+b)$ and $(a-b)$ both odd. $$\eqalign{ f(a + b) - f(a - b) = {(a + b)^2 - 1\over 4} - {(a - b)^2 - 1\over 4} \cr = ab.}$$ Case II $(a+b)$ and $(a-b)$ both even. $$\eqalign{ f(a + b) - f(a - b) = {(a + b)^2\over 4} - {(a - b)^2\over 4} \cr = ab.}$$