If 1+2+3+...+n=?
s = 1+2+3+ 1...+(n- 1)+n
S=n+(n- 1)+...+3+2+ 1
Then 2s = (n+1)+(n+1)+...+(n+1)+(n+1) n.
=n(n+ 1)
So S=n(n+ 1)/2.
Example 2
Find the sum of the first 2n terms of 2462n.
Answer:
2462n
2n2(n- 1)2(n-2)2
Let the sum of the first n terms be s, and the above two expressions are added.
2s =[2+(2n)]+[4+2(n- 1)]+[6+2(n-2)]++[(2n)+2]* * n 2n+2。
Therefore: S=n(2n+2)/2=n(n+ 1).