Use c=13, positive shift s=2, and n=13 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.
G001 — Use c=13, positive shift s=2, and n=13 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.
Definition and scope
Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate.
original c=13;shift=2;n=13;terms c/(shift+k)-c/(shift+k+1),k0..n-1
After
ALL interior terms cancel;S_n=c/shift-c/(shift+n)=169/30;signed remainder=13/15;reverse sum=13/2
Conditions: none
Every interior positive endpoint cancels its preceding negative occurrence,leaving the first endpoint and final negative one.Add the signed remainder back to reverse-check;do not replace a negative-c remainder by its absolute value.
same source formatting c=13;source remainder ratio text=13/15;source infinite-sum ratio text=13/2
After
computed partial sum REDUCED=169/30;remainder/infinite fields retain DIRECT UNREDUCED source text;numeric equivalents=[13/15, 13/2]
Conditions: none
The source reduces the computed partial sum but directly formats the other two ratio fields,which can remain2/4 or0/den.Preserve those answer bytes and verify their numerical values separately;mathematical equivalence does not authorize rewriting the source output.
DECLARED whole telescoping tail;remainder at general n is(13)/(n+2)
After
signed remainder tends to0;full partial-sum limit=c/shift under the given index convention
Conditions: none
The fixed coefficient and positive shift give an explicit remainder for every n.Its limit proves the full sum,not success of the single supplied endpoint check.
The full substitutions and premise checks give the original answer; source identity,format and status are unchanged.
This question review and source binding
Review date: 2026-10-26
Original questions, answers, step expressions, numeric ASTs and historical records are retained. Source candidate labels remain historical and are not overwritten as formal proof.