First term × number of terms+number of terms (number of terms-1)× tolerance /2.
{2 First item+(number of items-1)× number of tolerance items }/2.
n = 100x( 1+0.05)^n。
Sn = a 1+a2+...+ An.
= 100 x( 1+0.05)x[( 1+0.05)^n- 1]/[( 1+0.05)- 1].
= 2 100 x[( 1+0.05)^n- 1)].
In n years, what is the total?
=Sn .
= 2 100 x[( 1+0.05)^n- 1)].
This series is called arithmetic progression, and this constant is called arithmetic progression's tolerance, which is usually expressed by the letter D, for example: 1, 3,5,7,9 ... (2n-1).
The general formula of arithmetic progression {an} is: an = a1+(n-1) D. The first n terms and formulas are: sn = n * a1+n (n-1) d/2 or sn = n (a Note: All the above n are positive integers.
Other inferences:
① Sum = (first item+last item) × number of items ÷2.
② Number of items = (last item-first item) ÷ tolerance+1.
③ The first term =2x and the number of terms-the last term or the last term-tolerance × (the number of terms-1).
④ The last item =2x and the number of items-the first item.
⑤ The last term = the first term+(number of terms-1)× tolerance.
⑥2 (sum of the first 2n terms and-the first n terms) = sum of the first n terms and+the first 3n terms and-the first 2n terms.