Question

Find the sum: 1+2+3+ ... +9999+10000.

Answer 1

This requires the formula for summation of an arithmetic sequence, which was first derived by Gauss, since the sequence is of the form $u_{n} = n$.

a = 1 (the first term)
L = 10000 (the last term)
n = 10000 (the number of terms summed)

The formula is:
$S = \frac{1}{2}n(a + L)$
$S = ( \frac{1}{2}*10000)(1 + 10000) = 50005000$

Therefore the sum is 50005000

I hope this helps

Answer 2

It's an arithmetic sequence.

$1,2,3,...,9999,10000 \\ \\ a_1=1 \\ d=a_2-a_1=2-1=1$

Find which term is 10000:
$a_n=a_1+d(n-1) \\ 10 000=1+1(n-1) \\ 10000=1+n-1 \\ n=10000$
10000 is the 10000th term.

$S_n=\frac{n(a_1+a_n)}{2} \\ \Downarrow \\ S_{10 000}=\frac{10 000(1+10 000)}{2}=\fra{10 000 \times 10 001}{2}=5 000 \times 10 001=50005000 \\ \\ \boxed{1+2+3+...+9999+10000=50005000}$