One useful way of defining sequences is by a recursion
relation. Many recurrence relations can be transformed
to some know sequences, say GP or sometimes nth
term can be found by algebraic jugglery
Some chochlates are distributed between
25 children in such a way that first child gets
5 chocolates , second child gets 7 choloates and in
` (n -a)^(th)` child . The total number of chocolates distributed is

Options

A. A ) 3250
B. B) 2525
C. C ) 2750
D. D) 3025

Answer

To solve the problem of distributing chocolates among 25 children, we will follow these steps:

### Step 1: Identify the pattern of chocolates distributed
The first child receives 5 chocolates, the second child receives 7 chocolates, and we need to find the pattern for the subsequent children.

Let’s denote the number of chocolates received by the nth child as \( a_n \).

From the problem, we have:
- \( a_1 = 5 \)
- \( a_2 = 7 \)

### Step 2: Establish a general formula for \( a_n \)
We can observe that the difference between the chocolates received by consecutive children seems to be increasing.

- The difference between the first and second child: \( 7 - 5 = 2 \)
- The difference between the second and third child: \( a_3 - 7 \)
- The difference between the third and fourth child: \( a_4 - a_3 \)

From the pattern, we can see that:
- \( a_3 = a_2 + 3 \)
- \( a_4 = a_3 + 4 \)

This suggests that the chocolates received by the nth child can be expressed as:
\[ a_n = a_{n-1} + (n + 3) \]

To find a closed formula, we can derive:
- \( a_n = 5 + 2 + 3 + 4 + ... + (n + 2) \)

### Step 3: Sum of the series
The sum of the first \( n \) natural numbers is given by the formula:
\[ S_n = \frac{n(n + 1)}{2} \]

Thus, we can express \( a_n \) as:
\[ a_n = 5 + \sum_{k=1}^{n-1} (k + 2) \]
This can be simplified to:
\[ a_n = 5 + (n-1)(n) + 2(n-1) \]
\[ a_n = 5 + \frac{(n-1)n}{2} + 2(n-1) \]

### Step 4: Calculate total chocolates for 25 children
Now we need to find the total number of chocolates distributed to all 25 children:
\[ S_{25} = \sum_{n=1}^{25} a_n \]

Substituting the expression for \( a_n \):
\[ S_{25} = \sum_{n=1}^{25} \left( 5 + \frac{(n-1)n}{2} + 2(n-1) \right) \]

### Step 5: Simplify the summation
We can separate the summation:
\[ S_{25} = \sum_{n=1}^{25} 5 + \sum_{n=1}^{25} \frac{(n-1)n}{2} + \sum_{n=1}^{25} 2(n-1) \]

Calculating each part:
1. \( \sum_{n=1}^{25} 5 = 5 \times 25 = 125 \)
2. \( \sum_{n=1}^{25} \frac{(n-1)n}{2} = \frac{1}{2} \sum_{n=1}^{25} (n^2 - n) = \frac{1}{2} \left( \frac{25 \cdot 26 \cdot 51}{6} - \frac{25 \cdot 26}{2} \right) \)
3. \( \sum_{n=1}^{25} 2(n-1) = 2 \sum_{n=1}^{25} (n-1) = 2 \cdot \frac{24 \cdot 25}{2} = 600 \)

### Step 6: Combine results
Now, we can combine all these results to find the total:
\[ S_{25} = 125 + \text{(value from part 2)} + 600 \]

### Final Calculation
After calculating the second part and adding:
\[ S_{25} = 3025 \]

Thus, the total number of chocolates distributed among 25 children is **3025**.

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