Fill in the blanks:
(i) `sum_(k=1)^(n)k=1+2+3+……………+n=`…………
(ii) `sum_(k=1)^(n)k^(2)=1^(2)+2^(2)+3^(2)+………………….+n^(2)=`………………..
(iii) `sum_(k=1)^(n)k^(3)=1^(3)+2^(3)+3^(3)+…………..+n^(3)=`…………………

Answer

To solve the given question, we need to fill in the blanks for three summation formulas. Let's go through each part step by step.

### Part (i)
We need to find the sum of the first \( n \) natural numbers:
\[ \sum_{k=1}^{n} k = 1 + 2 + 3 + \ldots + n \]
The formula for the sum of the first \( n \) natural numbers is:
\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]
Thus, we fill in the blank:
\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]

### Part (ii)
Next, we need to find the sum of the squares of the first \( n \) natural numbers:
\[ \sum_{k=1}^{n} k^2 = 1^2 + 2^2 + 3^2 + \ldots + n^2 \]
The formula for the sum of the squares of the first \( n \) natural numbers is:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]
Thus, we fill in the blank:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]

### Part (iii)
Finally, we need to find the sum of the cubes of the first \( n \) natural numbers:
\[ \sum_{k=1}^{n} k^3 = 1^3 + 2^3 + 3^3 + \ldots + n^3 \]
The formula for the sum of the cubes of the first \( n \) natural numbers is:
\[ \sum_{k=1}^{n} k^3 = \left( \frac{n(n + 1)}{2} \right)^2 \]
Thus, we fill in the blank:
\[ \sum_{k=1}^{n} k^3 = \left( \frac{n(n + 1)}{2} \right)^2 \]

### Final Answers
1. \(\sum_{k=1}^{n} k = \frac{n(n + 1)}{2}\)
2. \(\sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6}\)
3. \(\sum_{k=1}^{n} k^3 = \left( \frac{n(n + 1)}{2} \right)^2\)