Question

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Answer 1

Find the sum starting from the n=1 to 12, of 9(4^n)-1

Answer 2

if n=1, 9(4^1)-1 = 36-1 = 35
if n=2, 9(4^2)-1 = 9(16)-1 = 143

the sum is 35+143+....sum where n=12

Answer 3

ok well, same thing. 9(4^(n-1)).

Let n=1.

9(4^(1-1) = 9(4^0) = 9(1) = 9.

Just add the sums from n=1 to 12, or use the shortcut.

Answer 4

The nth term, I would presume. S(n) is still sum, but it is used in partial sums, I think

Answer 5

It's the formula used to find nth partial sum of an Arithmetic series (the first S(n) forumla you posted)

Answer 6

Like if the following was used:
\[\sum_{n=1}^{10000}2n+6\]

It would take you forever to add up all the sums manually, That's a shortcut

Answer 7

Been a while since I did sums, so I forgot until I remembered. And you're welcome

Answer 8

The answer lies within your heart.