Question

Вычислите сумму 3 членов геометрической прогрессии b1=5 q=4 срочноо

Answer 1

Ответ:

S3=b1(1-q^3)/1-q

S3=5(1-4^3)/1-4

5(-63)/(-3)

S=5•21

S=105

Answer 2

Calculating the Sum of a Geometric Progression

To calculate the sum of the first three terms of a geometric progression with \( b_1 = 5 \) and \( q = 4 \), we can use the formula for the sum of the first \( n \) terms of a geometric progression:

Sum of the first \( n \) terms of a geometric progression: \[ S_n = \frac{{b_1 \cdot (q^n - 1)}}{{q - 1}} \]

Substituting the given values, we get: \[ S_3 = \frac{{5 \cdot (4^3 - 1)}}{{4 - 1}} \]

Calculating the value: \[ S_3 = \frac{{5 \cdot (64 - 1)}}{{3}} = \frac{{5 \cdot 63}}{{3}} = 105 \]

Therefore, the sum of the first three terms of the geometric progression is 105.