Question

Calculati sumele:
a) 1+5+9+13+...+93;
b) 6+16+26+36+...+136;
c) 3+10+17+24+...+178;
d) 1+6+11+16+...+146;
e) (-25)+(-20)+(-15)+(-10)+...+60

Va rog mult sa ma ajutati!!!

Answer 1

a) $1+5+9+...+93=\sum_{k=0}^{23}(1+4k)=\sum_{k=0}^{23}1+4\sum_{k=0}^{23}k$$=24+4\frac{23\cdot 24}{2}=24+4\cdot23\cdot12=1128$
b) $6+16+...+136=\sum_{k=0}^{13}(6+10k)=6\cdot14+10\sum_{k=0}^{13}k$$=84+10\cdot \frac{13\cdot14}{2}=994$
c) $3+10+17+...+178=\sum_{k=0}^{25}(3+7k)=3\cdot26+7\sum_{k=0}^{25}k$$=78+7\cdot\frac{25\cdot26}{2}=2353$
d) $1+6+11+...+146=\sum_{k=0}^{25}(1+5k)=26+5\sum_{k=0}^{25}k=26+5\cdot\frac{25\cdot26}{2}$$=1651$
e) $(-25)+(-20)+...+60=30+35+...+60=5+10+15+...+60$$-(5+10+...+25)$$=\sum_{k=1}^{12}(5k)-75=5\sum_{k=1}^{12}k-75=5\cdot\frac{12\cdot13}{2}-75=315$
Posibil sa fi gresit la calcul pe undeva, n-am stat sa verific.