Let `S_k` be sum of an indinite G.P whose first term is 'K' and commmon ratio is `(1)/(k+1)`. Then `Sigma_(k=1)^(10) S_k` is equal to _________.

Explanation

To solve the problem, we need to find the sum of an infinite geometric progression (G.P.) for each value of \( k \) from 1 to 10, and then sum these results.

### Step 1: Identify the formula for the sum of an infinite G.P.
The sum \( S \) of an infinite G.P. with first term \( a \) and common ratio \( r \) (where \( |r| < 1 \)) is given by:
\[ S = \frac{a}{1 - r} \]

### Step 2: Substitute the values for \( S_k \)
In our case, the first term \( a = k \) and the common ratio \( r = \frac{1}{k+1} \). Therefore, we can write:
\[ S_k = \frac{k}{1 - \frac{1}{k+1}} \]

### Step 3: Simplify the expression for \( S_k \)
Now, simplify the denominator:
\[ 1 - \frac{1}{k+1} = \frac{(k+1) - 1}{k+1} = \frac{k}{k+1} \]
Thus, we can rewrite \( S_k \):
\[ S_k = \frac{k}{\frac{k}{k+1}} = k \cdot \frac{k+1}{k} = k + 1 \]

### Step 4: Find the summation \( \Sigma_{k=1}^{10} S_k \)
Now we need to find:
\[ \Sigma_{k=1}^{10} S_k = \Sigma_{k=1}^{10} (k + 1) \]
This can be separated into two sums:
\[ \Sigma_{k=1}^{10} S_k = \Sigma_{k=1}^{10} k + \Sigma_{k=1}^{10} 1 \]

### Step 5: Calculate \( \Sigma_{k=1}^{10} k \)
The sum of the first \( n \) natural numbers is given by the formula:
\[ \Sigma_{k=1}^{n} k = \frac{n(n + 1)}{2} \]
For \( n = 10 \):
\[ \Sigma_{k=1}^{10} k = \frac{10 \times 11}{2} = 55 \]

### Step 6: Calculate \( \Sigma_{k=1}^{10} 1 \)
This sum simply counts the number of terms, which is 10:
\[ \Sigma_{k=1}^{10} 1 = 10 \]

### Step 7: Combine the results
Now combine the two sums:
\[ \Sigma_{k=1}^{10} S_k = 55 + 10 = 65 \]

### Final Answer
Thus, the value of \( \Sigma_{k=1}^{10} S_k \) is:
\[ \boxed{65} \]