Question

1-3+3^2-3^3+...+3^6 Help please

Answer 1

1-3+3²-3³+...+3⁶

3⁰-3¹+3²-3³+...+3⁶

Геометрическая прогрессия

b1 = 1

q = -3

S7 = 1-3+3²-3³+...3⁶ = b1(q⁷-1)/q-1 = 3⁷-1/3-1 = 3⁷-1/2 = -2187-1/2 = -2188/2 = -1092

Answer 2

It looks like you want to calculate the sum of a series that alternates between addition and subtraction of powers of 3. The series you've provided is:

1 - 3 + 3^2 - 3^3 + ... + 3^6

To calculate the sum, we can use the formula for the sum of a finite geometric series:

Sum = a * (1 - r^n) / (1 - r)

In this formula:

• "a" is the first term of the series.
• "r" is the common ratio between successive terms.
• "n" is the number of terms.

In your series, the first term "a" is 1, the common ratio "r" is -3 (since the sign alternates), and there are 7 terms (from 3^0 to 3^6).

Let's plug these values into the formula:

Sum = 1 * (1 - (-3)^7) / (1 - (-3))

Simplify the exponents:

Sum = 1 * (1 - (-2187)) / (1 + 3)

Sum = 1 * (1 + 2187) / 4

Sum = 2188 / 4

Sum = 547

So, the sum of the series is 547.