Răspuns:
S1=∑j=1+2+...+i=i(i+1)/2=
(i²+i)/2
S2=∑(i²+i)/2=
∑i²/2+∑i/2
1/2∑i²=1/2(1²+2²+...+2013²)=
1/2(2013*2014*(2*2013+1)/6=
(2013*2014*4017)/12=...
1/2∑i=1/2(1+2+...+2013)=1/2*2013*2014/2=
2013*2014/4
S2=(2013*2014*2017)/12+2013*2014/4=
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