1 year ago

Find the 9th partial sum of the geometric sequence an=43n–1.

Mathematics

1 answer:

siniylev [52]1 year ago

7 0

Answer:

39364.

Step-by-step explanation:

I guess its an = 4*(3)^(n-1)

r = 3, a1 = 4 and n = 9

Sum of n terms = a1 * (r^n - 1) / (r - 1)

so S9 = 4 *(3^9 - 1) / (3 - 1)

= 4 * (19683 - 1) / 2

= 78728/2

= 39364.

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