# prove that the sum of the integers is

• Jan 14th 2010, 03:30 AM
differentiate
prove that the sum of the integers is
The positive integers are bracketed as follows:
(1), (2,3), (4,5,6), . . .
where there are r integers in the rth bracket. Prove that the sum of the integers in the rth bracket is

$\displaystyle \frac{r}{2} (r^2 +1)$

this is what I did
using the formula for arithmetic sequence,
we know that there are r integers. and the difference is 1

so that becomes
Sum of r integers = $\displaystyle \frac{r}{2}(2a + (r-1))$ where a is the first term of the sequence

However, I can't figure out what a (a.k.a the first term) is equal to

any ideas?

thanks
• Jan 14th 2010, 04:47 AM
Laurent
Quote:

Originally Posted by differentiate
The positive integers are bracketed as follows:
(1), (2,3), (4,5,6), . . .
where there are r integers in the rth bracket. Prove that the sum of the integers in the rth bracket is

$\displaystyle \frac{r}{2} (r^2 +1)$

this is what I did
using the formula for arithmetic sequence,
we know that there are r integers. and the difference is 1

so that becomes
Sum of r integers = $\displaystyle \frac{r}{2}(2a + (r-1))$ where a is the first term of the sequence

However, I can't figure out what a (a.k.a the first term) is equal to

Note that $\displaystyle a=(1+2+3+4+\cdots+(r-1))+1$ because $\displaystyle a$ is comes after the first bracket (1 term), the second brack (2 terms), etc. , and after the $\displaystyle (r-1)$-th bracket.
• Jan 14th 2010, 04:59 AM
differentiate
ohh that makes sense now

thanks!