Question

Bn=128,q=2,n=7.Найдите b1 и sn

Answer 1

Bn=b1×q^n-1; b6=b1×q^6; 128=b1×2^6;
128=128×b1; b1=1;
S7=1×(2^7-1)/2-1= 128-1=127;
ответ: b1=1; S7=127.

Answer 2

Формула для n-ного члена геометрической прогрессии
$b_{n} =b _{1} * q^{n-1} ,n=7, [tex] b_{n} =128$
Выражаем $b_{1} = \frac{ b_{7} }{q ^{6} }$
Отсюда получаем $b_{1} =2$
$s_{n} = \frac{ b_{1}*( q^{n}-1) }{q-1}$
$s_{n} =254$

Answer 3

To find the values of b₁ and Sₙ, we need to understand the given notation. It appears that Bₙ represents the base of a number system, q represents the quotient, and n represents the exponent.

In this case, Bₙ is given as 128, q is given as 2, and n is given as 7.

To find b₁, we can substitute n = 1 in the formula:

b₁ = Bₙ / (q^(n-1))

b₁ = 128 / (2^(1-1)) b₁ = 128 / 2^0 b₁ = 128 / 1 b₁ = 128

Therefore, b₁ is equal to 128.

To find Sₙ, we can use the formula:

Sₙ = (b₁ * (q^n - 1)) / (q - 1)

Sₙ = (128 * (2^7 - 1)) / (2 - 1) Sₙ = (128 * (128 - 1)) / 1 Sₙ = (128 * 127) / 1 Sₙ = 16256

Therefore, Sₙ is equal to 16256.

To summarize: b₁ = 128 Sₙ = 16256