1,a(1) = a,a(n)為公差為r的等差數(shù)列.
1-1,通項公式,
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.
可用歸納法證明.
n = 1 時,a(1) = a + (1-1)r = a.成立.
假設 n = k 時,等差數(shù)列的通項公式成立.a(k) = a + (k-1)r
則,n = k+1時,a(k+1) = a(k) + r = a + (k-1)r + r = a + [(k+1) - 1]r.
通項公式也成立.
因此,由歸納法知,等差數(shù)列的通項公式是正確的.
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
同樣,可用歸納法證明求和公式.（略）
2,a(1) = a,a(n)為公比為r（r不等于0）的等比數(shù)列.
2-1,通項公式,
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).
可用歸納法證明等比數(shù)列的通項公式.（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時,
S(n) = a[1 - r^n]/[1-r]
r = 1時,
S(n) = na.
同樣,可用歸納法證明求和公式.（略）