(1) an equal proportion and an arithmetic difference: the sum of arithmetic differences must be =n(n+ 1)/2, and the geometric series needs to be divided into cases:

When a= 1 and = n；; When a≠ 1, and = a (1-a)/(1-a) according to the summation formula of equal ratio;

So the sum of the original series is also divided into two situations:

When a= 1 and = n-n (n+1)/2 = n (1-n)/2;

When a≠ 1, and = a (1-an n)/(1-a)-n (n+1)/2;

(2) Equal ratio and arithmetic difference:

Sum of arithmetic = n (2+2n)/2 = n (n+1); Proportional sum = 3 [1-5 (-n)]/5 (1-kloc-0//5) = 3 [1-5 (-n)]/4,

So the sum of the original sequences = n (n+1)-3 [1-5 (-n)]/4;

(3) Classification discussion:

When x=0 and = 1;

When x= 1 and = n (n+ 1)/2;

When x≠0 or 1, use the dislocation elimination method (the fixed method of proportional arithmetic multiplication or proportional arithmetic division must be mastered):

Let the sum be Sn,

Sn =1+2x+3x2+...............+(n-1) x (n-2)+nx2 (n-1), ①, multiply both sides of this formula by the common ratio of geometric series.

xSn=x+2x^2+.......................................(n- 1)x^(n- 1)+nx^n,②

①-②De:( 1-x)Sn = 1+[x+x2+...+x (n- 1)]-NX n

= 1+(x^n-x)/(x- 1)-nx^n

So sn = (1-nxn)/* (1-x)-(xn-x)/(1-x) 2.

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