Sum of the series 1+(1+2)+(1+2+3)+(1+2+3+4)+... to n terms be `1/m n(n+1)[(2n+1)/k+1]` then find the value of k+m

Answer

To find the sum of the series \( S = 1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + \ldots \) up to \( n \) terms, we can follow these steps:

### Step 1: Understand the series
The series can be rewritten as:
\[ S = 1 + (1 + 2) + (1 + 2 + 3) + \ldots + (1 + 2 + \ldots + n) \]
Each term in the series is the sum of the first \( k \) natural numbers, where \( k \) ranges from 1 to \( n \).

### Step 2: Use the formula for the sum of the first \( k \) natural numbers
The sum of the first \( k \) natural numbers is given by:
\[ \text{Sum} = \frac{k(k + 1)}{2} \]
Thus, we can express the series \( S \) as:
\[ S = \sum_{k=1}^{n} \frac{k(k + 1)}{2} \]

### Step 3: Factor out the constant
Factoring out \( \frac{1}{2} \) from the summation:
\[ S = \frac{1}{2} \sum_{k=1}^{n} k(k + 1) \]

### Step 4: Expand the summation
We can expand \( k(k + 1) \):
\[ k(k + 1) = k^2 + k \]
Thus,
\[ S = \frac{1}{2} \left( \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k \right) \]

### Step 5: Use the formulas for the sums
We know the formulas for the sums:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]
\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]
Substituting these into our expression for \( S \):
\[ S = \frac{1}{2} \left( \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2} \right) \]

### Step 6: Combine the terms
To combine these fractions, we need a common denominator. The common denominator is 6:
\[ S = \frac{1}{2} \left( \frac{n(n + 1)(2n + 1)}{6} + \frac{3n(n + 1)}{6} \right) \]
This simplifies to:
\[ S = \frac{1}{2} \cdot \frac{n(n + 1)(2n + 1 + 3)}{6} = \frac{1}{2} \cdot \frac{n(n + 1)(2n + 4)}{6} \]
\[ = \frac{n(n + 1)(n + 2)}{6} \]

### Step 7: Final expression for \( S \)
Thus, we have:
\[ S = \frac{n(n + 1)(n + 2)}{6} \]

### Step 8: Compare with the given expression
We are given that:
\[ S = \frac{1}{m} n(n + 1) \left( \frac{2n + 1}{k + 1} \right) \]
Setting the two expressions equal:
\[ \frac{n(n + 1)(n + 2)}{6} = \frac{1}{m} n(n + 1) \left( \frac{2n + 1}{k + 1} \right) \]

### Step 9: Simplify and find \( m \) and \( k \)
Cancelling \( n(n + 1) \) from both sides (assuming \( n \neq 0 \)):
\[ \frac{n + 2}{6} = \frac{1}{m} \cdot \frac{2n + 1}{k + 1} \]
This leads to:
\[ m(n + 2) = \frac{6(2n + 1)}{k + 1} \]

### Step 10: Determine values of \( m \) and \( k \)
By comparing coefficients, we can find:
- Let \( m = 4 \)
- Let \( k = 3 \)

### Step 11: Calculate \( k + m \)
Thus:
\[ k + m = 3 + 4 = 7 \]

### Final Answer
The value of \( k + m \) is \( 7 \).

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