The first n terms and formulas are: Sn=na 1+n(n- 1)d/2 or Sn=n(a 1+an)/2 (n is a natural number).

A 1 is the first term, an is the last term, n is the number of terms, and d is the tolerance of arithmetic progression.

Geometric series an = a1× q (n-1);

sum:sn = a 1( 1-q n)/( 1-q)=(a 1-an×q)/( 1-q)(q≠ 1。

The way to deduce the sum formula of the first n terms of arithmetic progression is to arrange a series in reverse order (reverse order), and then add it to the original series to get n (a 1+an).

Sn？ = A1+A2+A3+...+An

Sn？ =an+ an- 1+an-2......+a 1

Add up and down to get Sn=(a 1+an)n/2.

Extended data:

To prove a proposition related to a positive integer n, there are the following steps:

(1) proves that the proposition holds when n takes the first value;

(2) Assuming that n = k (the first value of k ≥ n, and k is a natural number), it is proved that the proposition is also true when n=k+ 1.

Example:

Verification:

1×2×3×4 + 2×3×4×5 + 3×4×5×6 + .……+n(n+ 1)(n+2)(n+3)=[n(n+ 1)(n+2)(n+3)(n+4)]/5

Prove:

When n= 1, there are:

1×2×3×4 = 24 = 2×3×4×5/5

Assuming that the proposition holds when n=k, then:

1×2×3×4+2×3×4×5+3×4×5×6+。 ……+k(k+ 1)(k+2)(k+3)=[k(k+ 1)(k+2)(k+3)(k+4)]/5

Then when n=k+ 1, there is:

1×2×3×4+2×3×4×5+3×4×5×6+……+(k+ 1)(k+2)(k+3)(k+4)

= 1×2×3×4+2×3×4×5+3×4×5×6+……+k(k+ 1)(k+2)(k+3)+(k+ 1)(k+2)(k+3)(k+4)

=[k(k+ 1)(k+2)(k+3)(k+4)]/5+(k+ 1)(k+2)(k+3)(k+4)

=(k+ 1)(k+2)(k+3)(k+4)*(k/5+ 1)

=[(k+ 1)(k+2)(k+3)(k+4)(k+5)]/5

That is, when n=k+ 1, the original equation still holds, which is proved by induction.

Baidu Encyclopedia-Sum of Series