Find the sum to n terms: `5.6+6.7+7.8+………………`

Answer

To find the sum to n terms of the sequence \(5.6 + 6.7 + 7.8 + \ldots\), we can represent the \(n\)-th term of the sequence in a more manageable form.

### Step 1: Identify the \(n\)-th term
The first term is \(5.6\), the second term is \(6.7\), and the third term is \(7.8\). We can see that the \(n\)-th term can be expressed as:
\[ t_n = (n + 4)(n + 5) \]
This is because:
- For \(n = 1\): \(t_1 = 5 \cdot 6\)
- For \(n = 2\): \(t_2 = 6 \cdot 7\)
- For \(n = 3\): \(t_3 = 7 \cdot 8\)

### Step 2: Write the sum to n terms
We need to find the sum:
\[ S_n = \sum_{r=1}^{n} t_r = \sum_{r=1}^{n} (r + 4)(r + 5) \]
Expanding this:
\[ t_r = (r + 4)(r + 5) = r^2 + 5r + 4r + 20 = r^2 + 9r + 20 \]
Thus, the sum becomes:
\[ S_n = \sum_{r=1}^{n} (r^2 + 9r + 20) \]

### Step 3: Separate the summation
We can separate the summation into three parts:
\[ S_n = \sum_{r=1}^{n} r^2 + \sum_{r=1}^{n} 9r + \sum_{r=1}^{n} 20 \]
This can be simplified as:
\[ S_n = \sum_{r=1}^{n} r^2 + 9\sum_{r=1}^{n} r + 20\sum_{r=1}^{n} 1 \]

### Step 4: Use summation formulas
We can use the formulas for the sums:
1. The sum of the first \(n\) natural numbers:
\[ \sum_{r=1}^{n} r = \frac{n(n + 1)}{2} \]
2. The sum of the squares of the first \(n\) natural numbers:
\[ \sum_{r=1}^{n} r^2 = \frac{n(n + 1)(2n + 1)}{6} \]
3. The sum of \(1\) for \(n\) terms is simply \(n\).

Substituting these into our equation:
\[ S_n = \frac{n(n + 1)(2n + 1)}{6} + 9 \cdot \frac{n(n + 1)}{2} + 20n \]

### Step 5: Simplify the expression
Now we can simplify:
\[ S_n = \frac{n(n + 1)(2n + 1)}{6} + \frac{9n(n + 1)}{2} + 20n \]
To combine these, we need a common denominator, which is \(6\):
\[ S_n = \frac{n(n + 1)(2n + 1)}{6} + \frac{27n(n + 1)}{6} + \frac{120n}{6} \]
Combining the fractions:
\[ S_n = \frac{n(n + 1)(2n + 1) + 27n(n + 1) + 120n}{6} \]

### Step 6: Factor and simplify further
Now we can factor out \(n\) from the numerator:
\[ S_n = \frac{n \left[(n + 1)(2n + 1) + 27(n + 1) + 120\right]}{6} \]
This can be simplified further, but the expression is already in a manageable form.

### Final Result
Thus, the sum of the first \(n\) terms of the sequence is:
\[ S_n = \frac{n(n + 1)(2n + 1 + 27) + 120n}{6} \]