Question

calculati: a) (5 la puterea 1+2+3 +....+20 + 4×5 la puterea 210)÷ 5 la puterea 211; b) [(1+3+5+...+19)÷(2×5) la puterea 2]×2 la puterea 3 ; c) (11×3 + 11×3 la puterea 2 +...+ 11×3 la puterea 64)÷( 1+3+3 la puterea 2 +....+3la puterea 63). Va rog sa ma ajutati.Multumesc frumos.

Answer 1

a) (5¹⁺²⁺³⁺ ⁺²⁰+ 4×5²¹⁰)÷ 5²¹¹=(5²¹⁰+ 4×5²¹⁰)÷ 5²¹¹=5²¹⁰(1+4):5²¹¹=5²¹¹:5²¹¹=1
1+2+3+..+20=20*21/2=210


b) [(1+3+5+...+19)÷(2×5)²]×2³=
1+3+5+...+2n-1=n²
2n-1=19  n=10
=( 10²÷10²)×2³=
=1×2³=8

c) (11×3 +11×3² +...+ 11×3⁶⁴)÷( 1+3+3² +....+3⁶³)=
= 11×3(1 +3 +...+ 3⁶³)÷( 1+3+3² +....+3⁶³)=33

Answer 2

a)5la1+2+..20=5la(20x21):2=5la puterea 210

(5la 210+4x5la210):5la211=    (5la210 factor comun)
5la210(1+4):5la211=
5la210x5):5la211=
5la211:5la211=1

b)[(1+3+5+...19):(2x5)la2]x2la3=
[(1+1+2+2+3+3+4+ ....+9+10):10la2]x2la3=
[(1+2+3+...9)+(1+2+3+..10)):10la2]x2la3=
[(9x10):2+(10x11):2):10la2]x2la3=
[(45+55):100]x8=
(100:100)x8=8

c)(11x3+11x3la2+..11x3la64):(1+3+3la2+3a3+...3la63)=
11 factor comun in prima paranteza
11(3+3la2+3la3+..3la64):(1+3+3la2+..3la63)=
11(3+3la2+...3la64)+11(3+3la2+..3la64):(3+3la2+3la3+..3la63)=
11(3+3la2+..3la64)+11x3=
11(3+3la2+..3la64)+33

nu sunt sigura de ultimul daca e ok