Question

calculati:
a) 1+2+2²+2³+....2 la puterea 100
b) 2×(1+3+3²+....3 la puterea49)
c) 4×(1+5+5²+....+5 la puterea 100)
d) 5×(6+6²+6³+....+6la puterea1000)

Answer 1

a) Progresie geometrica cu ratia q=2, b(1)=1   b(101)=2 la puterea 100
Formula: Sn=$b1* \frac{q^{n}-1 }{q-1}$

1+2+...+2 la puterea 100 = $1* \frac{2^{101}-1}{2-1} =2^{101}-1$

b) Progresie geometrica cu ratia q=3 , b(1)=1  b(50)= 3 la puterea 49

1+3+...+3 la puterea 49 = $1* \frac{3^{50}-1}{3-1} = \frac{3^{50}-1}{2}$
2*(1+3+...+ 3 la puterea 49) = $2* \frac{3^{50}-1}{2} =3^{50}-1$

c) Progresie geometrica cu ratia q=5 , b(1)=1 , b(101)=5 la puterea 100

1+5+...+5 la 100 = $1* \frac{5^{101}-1}{5-1} = \frac{5^{101}-1}{4}$
4*(1+5+...+5 la 100) = $4* \frac{5^{101}-1}{4} =5^{101}-1$

d) Te las pe tine:)