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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