Question

indentificand un factor comun, in fiecare suma, calculati:
a) 10+20=30+...+1230
b) 7+14+21+...+315
VA OG ESTE FOARTE URGENT, OFER COROANA

Answer 1

$\displaystyle a).10+20+30+...+1230=10(1+2+3+...+123)= \\ \\ =10\cdot \frac{123(123+1)}{2} =10 \cdot \frac{123 \cdot 124}{2} =10 \cdot \frac{15252}{2} =10 \cdot 7626=76260 \\ \\ b).7+14+21+...+315=7(1+2+3+...+45)= \\ \\ =7 \cdot \frac{45(45+1)}{2} =7 \cdot \frac{45 \cdot 46}{2} =7 \cdot \frac{2070}{2} =7 \cdot 1035=7245$

Answer 2

a.). 10*(1+2+3+...+123)=10[n(n+1):2] =10(123*124:2)=10*(123*62)=10*7526=75260 b.) 7(1+2+3+...+47)=7(47*48:2)=7(47*24)=7*1128=7896.