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