1,a(1) = a,a(n)为公差为r的等差数列.

1-1,通项公式,

a(n) = a(n-1) + r = a(n-2) + 2r = ...= a[n-(n-1)] + (n-1)r = a(1) + (n-1)r = a + (n-1)r.

可用归纳法证明.

n = 1 时,a(1) = a + (1-1)r = a.成立.

假设 n = k 时,等差数列的通项公式成立.a(k) = a + (k-1)r

则,n = k+1时,a(k+1) = a(k) + r = a + (k-1)r + r = a + [(k+1) - 1]r.

通项公式也成立.

因此,由归纳法知,等差数列的通项公式是正确的.

1-2,求和公式,

S(n) = a(1) + a(2) + ...+ a(n)

= a + (a + r) + ...+ [a + (n-1)r]

= na + r[1 + 2 + ...+ (n-1)]

= na + n(n-1)r/2

同样,可用归纳法证明求和公式.（略）

2,a(1) = a,a(n)为公比为r（r不等于0）的等比数列.

2-1,通项公式,

a(n) = a(n-1)r = a(n-2)r^2 = ...= a[n-(n-1)]r^(n-1) = a(1)r^(n-1) = ar^(n-1).

可用归纳法证明等比数列的通项公式.（略）

2-2,求和公式,

S(n) = a(1) + a(2) + ...+ a(n)

= a + ar + ...+ ar^(n-1)

= a[1 + r + ...+ r^(n-1)]

r 不等于 1时,

S(n) = a[1 - r^n]/[1-r]

r = 1时,

S(n) = na.

同样,可用归纳法证明求和公式.（略）