Current location - Training Enrollment Network - Mathematics courses - Another math problem of arithmetic geometric series.
Another math problem of arithmetic geometric series.
Answer: Choose D: It must be a geometric series, but it cannot be an arithmetic series. (excluding a= 1, ∫ If a= 1, then a? n? =0)

∵a? n+ 1? /a? n? =(S? n+ 1? -S? n? )/(S? n? -S? n- 1? )=[(a^(n+ 1)- 1)-(a^n- 1)]/[(a^n- 1)-a^(n- 1)- 1]

=[a(n+ 1)-a n]/[a n-a(n- 1)]=(a n)(a- 1)/[(a n)(65438+)

What about a? n+ 1? -a? n? =[S? n+ 1? -S? n? ]-[S? n? -S? n- 1? ]=S? n+ 1? -2S? n? +S? n- 1? =[a^(n+ 1)- 1]-2[a^n- 1]+[a^(n- 1)- 1]

=a^(n+ 1)-2a^n+a^(n- 1)=(a^n)[a-2+ 1/a]=(a^n)[(a? -2a+ 1)/a]=(a^n)(a- 1)? /a≠ constant, so it must not be arithmetic progression.

Generally speaking, the first n terms of geometric series and s? n? =a? (q? - 1)/(q- 1)=k(q? - 1)=k(a? -1), which is an exponential function;

When k= 1, the title form is obtained; And arithmetic progression's top N and S? n? =na? +n(n- 1)d/2=(d/2)n? +(a? -d/2)n=An? The form of +Bn is a quadratic function about the missing constant term of n.