Find the partial sum 1+3+9+…+2187 of a geometric sequence

Answer 1

To find the partial sum of a geometric sequence, we use the formula

\[S_n = a \left( \dfrac{1 - r^n}{1 - r} \right)\]

where:
- \( S_n \) is the partial sum of the sequence
- \( a \) is the first term of the sequence
- \( r \) is the common ratio between consecutive terms
- \( n \) is the number of terms in the sequence

In this case, the first term \( a = 1 \), the common ratio \( r = 3 \), and the last term we want to find the sum up to is \( 2187 \).

To find the number of terms, we can use the formula for the nth term of a geometric sequence:

\[a_n = a \cdot r^{n-1}\]

Since the last term is 2187, we have:

\[2187 = 1 \cdot 3^{n-1}\]

\[3^{n-1} = 2187\]

\[3^{n-1} = 3^7\]

Therefore, n - 1 = 7 and n = 8.

Now we can plug these values into the formula for the partial sum:

\[S_8 = 1 \left( \dfrac{1 - 3^8}{1 - 3} \right)\]

\[S_8 = 1 \left( \dfrac{1 - 6561}{-2} \right)\]

\[S_8 = 1 \left( \dfrac{-6560}{-2} \right)\]

\[S_8 = 1 \cdot 3280\]

\[S_8 = 3280\]

Therefore, the partial sum 1+3+9+…+2187 of a geometric sequence is 3280.