# How do you use summation notation to expression the sum # 5+15+45+...+3645#?

Then:

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To express the sum (5+15+45+...+3645) using summation notation, you can use the formula for the general term of the sequence and the summation symbol. The sequence appears to be a geometric sequence where each term is obtained by multiplying the previous term by 3 and adding 5.

The general term of the sequence can be represented as (a_n = 5 \times 3^{n-1}), where (n) is the position of the term in the sequence.

Thus, the sum can be expressed as:

[ \sum_{n=1}^{N} 5 \times 3^{n-1} ]

where (N) is the number of terms in the sequence. To find (N), you can use the formula for the nth term:

[a_n = 5 \times 3^{n-1}]

Setting (a_n = 3645) and solving for (n):

[3645 = 5 \times 3^{n-1}]

[3^{n-1} = \frac{3645}{5}]

[3^{n-1} = 729]

[n-1 = \log_3(729)]

[n-1 = 6]

[n = 7]

Thus, there are 7 terms in the sequence. Therefore, the summation notation for the given sequence is:

[ \sum_{n=1}^{7} 5 \times 3^{n-1} ]

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The given series is a geometric series where each term is multiplied by 3 to get the next term, starting from 5. This can be expressed as (5, 15, 45, \ldots, 3645), where the first term (a = 5) and the common ratio (r = 3).

The sum of a geometric series can be expressed using the formula:

[ S_n = \frac{a(r^n - 1)}{r - 1} ]

To find (n), the number of terms, we use the last term formula for a geometric series:

[ a \cdot r^{n-1} = 3645 ]

Substituting (a = 5) and (r = 3):

[ 5 \cdot 3^{n-1} = 3645 ]

[ 3^{n-1} = \frac{3645}{5} ]

[ 3^{n-1} = 729 ]

Since (729 = 3^6), we have:

[ 3^{n-1} = 3^6 ]

[ n-1 = 6 ]

[ n = 7 ]

Therefore, the sum of the series is the sum of the first 7 terms of this geometric sequence, which can be expressed in summation notation as:

[ S = \sum_{k=0}^{6} 5 \cdot 3^k ]

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