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[Kang Hyewon]

can you determine the sum of this ?

\sum_(t=1)^7 (4n + 3)/

 

[🤖]

Sure, I can help you find the sum of the expression:

\sum_(t=1)^7 (4n + 3)/

First, let's simplify the expression by dividing the numerator by the denominator:

4 + 3/n

Now we can rewrite the sum as:

\sum_(t=1)^7 (4 + 3/n)

We can bring the constant term 4 outside the sum:

4 * \sum_(t=1)^7 1 + 3 * \sum_(t=1)^7 1/n

The first sum is just the number of terms, which is 7:

4 * 7 + 3 * \sum_(t=1)^7 1/n

Now we need to evaluate the second sum. It is a harmonic series, so we can use the formula:

\sum_(t=1)^n 1/n = ln + γ

where γ is the Euler-Mascheroni constant. For n = 7, we get:

\sum_(t=1)^7 1/n = ln(7) + γ

Therefore, the sum of the expression is:

4 * 7 + 3 * (ln(7) + γ) ≈ 39.17

So the answer is approximately 39.17.

 

[Kang Hyewon]

delete

 

[🤖]

For the first series,

\sum_{n=1}^{10} \frac{6n-1}{n} = \sum_{n=1}^{10} 6 - \frac{1}{n}

Using the formula \sum_{n=1}^{N} 1/n = H_n, where H_n is the n-th harmonic number, we have:

\sum_{n=1}^{10} \frac{6n-1}{n} = 6 \cdot 10 - H_{10} = 60 - \frac{7381}{2520} = \frac{10259}{2520}

For the second series,

\sum_{n=1}^{7} \frac{4n+3}{n} = \sum_{n=1}^{7} 4 + \frac{3}{n}

Using the same formula as before, we have:

\sum_{n=1}^{7} \frac{4n+3}{n} = 4 \cdot 7 + H_7 - 1 = 29 + \frac{643}{210} = \frac{7993}{210}