Question

Faceți 21:b,c va rog foarte mult dau 15 puncte și coroana

Answer 1

$b)(4 + 8 + 12 + 16 + ... + 8000) \div (2 + 4 + 6 + 8 + ... + 4000) + ( {2}^{53} + {2}^{52} + {2}^{51} ) \div { ({2}^{17}) }^{3}$

$= [4(1 + 2 + 3 + 4 + ... + 2000)] \div [2(1 + 2 + 3 + 4 + ... + 2000)] + ( {2}^{53} + {2}^{52} + {2}^{51} ) \div {2}^{17 \times 3}$

$= (4 \div 2)[(1 + 2 + 3 + 4 + ... + 2000) \div (1 + 2 + 3 +4+ ... + 2000)] + ( {2}^{53} + {2}^{52} + {2}^{51} ) \div {2}^{51}$

$= 2 \times 1 + {2}^{53} \div {2}^{51} + {2}^{52} \div {2}^{51} + {2}^{51} \div {2}^{51}$

$= 2 + {2}^{53 - 51} + {2}^{52 - 51} + 1$

$= 2 + {2}^{2} + {2}^{1} + 2$

$= 2 + 4 + 2 + 2$

$= 10$

$c) {(6 + 12 + 18 + 24 + ... + 600)}^{2} \div (3 + 6 + 9 + 12 + ... + 300) \div (1 + 2 + 3 + ... + 100)$

$= {[6(1 + 2 + 3 + 4 + ... + 100)]}^{2} \div [3(1 + 2 + 3 + 4 + ... + 100)] \div (1 + 2 + 3 + ... + 100)$

$= ( {6}^{2} \div 3 \div 1)[ {(1 + 2 + 3 + 4 + ... + 100)}^{2} \div (1 + 2 + 3 + 4 + ... + 100) \div (1 + 2 + 3 + 4 + ... + 100) ]$

$= (36 \div 3) {(1 + 2 + 3 +4 + ...} + 100)^{2 - 1 - 1}$

$= 12 \times {(1 + 2 + 3 +4+ ... + 100)}^{0}$

$= 12 \times 1$

$=12$