Question

由等比数列和等差数列之积组成的数列怎样求通项公式

Answer 1

用交错相差法求。

a(n) = a+(n-1)d,
b(n) = bq^(n-1),

t(n) = a(1)b(1) + a(2)b(2) + a(3)b(3) + ... + a(n-1)b(n-1) + a(n)b(n)
= a*b + (a+d)bq + (a+2d)bq^2 + ... + [a+(n-2)d]bq^(n-2) + [a+(n-1)d]bq^(n-1),

q=1时，b(n) = b, t(n) = b[a(1)+a(2)+...+a(n)] = b[na + n(n-1)d/2] = nab + n(n-1)bd/2.
q不为1时，qt(n) = abq + (a+d)bq^2 + ... + [a+(n-2)d]bq^(n-1) + [a+(n-1)d]bq^n,

(1-q)t(n) = t(n) - qt(n) = ab + dbq + dbq^2 + ... + dbq^(n-1) - [a+(n-1)d]bq^n
= ab + dbq[1 + q + ... + q^(n-2)] - [a+(n-1)d]bq^n
= ab + dbq[1-q^(n-1)]/(1-q) - [a+(n-1)d]bq^n
= ab + dbq/(1-q) - [a + d/(1-q) + (n-1)d]bq^n,

t(n) = ab/(1-q) + dbq/(1-q)^2 - [a/(1-q)+d/(1-q)^2 + (n-1)d/(1-q)]bq^n.

综合，有，
q=1时，a(1)b(1)+a(2)b(2)+...+a(n)b(n) = nab + n(n-1)db/2.
q不为1时，a(1)b(1)+a(2)b(2)+...+a(n)b(n) = ab/(1-q) + dbq/(1-q)^2 - [a/(1-q) + d/(1-q)^2 + (n-1)d/(1-q)]bq^n.