Evaluate the Summation sum from n=51 to 100 of 5n

Solution:

Step 1:

Transform the original summation to start from $n=1$.

$$\sum_{n=51}^{100} 5n = \sum_{n=1}^{100} 5n - \sum_{n=1}^{50} 5n$$

Step 2:

Compute the summation from $n=1$ to $n=100$.

Step 2.1:

Extract the constant $5$ from the summation.

$$5 \sum_{n=1}^{100} n$$

Step 2.2:

Apply the arithmetic series sum formula.

$$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$$

Step 2.3:

Insert the upper limit into the formula and multiply by $5$.

$$5 \left( \frac{100(100+1)}{2} \right)$$

Step 2.4:

Carry out the simplification process.

Step 2.4.1:

Add together $100$ and $1$.

$$5 \times \frac{100 \times 101}{2}$$

Step 2.4.2:

Calculate $100 \times 101$.

$$5 \left( \frac{10100}{2} \right)$$

Step 2.4.3:

Divide $10100$ by $2$.

$$5 \times 5050$$

Step 2.4.4:

Multiply $5$ by $5050$.

$$25250$$

Step 3:

Compute the summation from $n=1$ to $n=50$.

Step 3.1:

Extract the constant $5$ from the summation.

$$5 \sum_{n=1}^{50} n$$

Step 3.2:

Use the arithmetic series sum formula.

$$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$$

Step 3.3:

Place the upper limit into the formula and multiply by $5$.

$$5 \left( \frac{50(50+1)}{2} \right)$$

Step 3.4:

Proceed with the simplification.

Step 3.4.1:

Combine $50$ and $1$.

$$5 \times \frac{50 \times 51}{2}$$

Step 3.4.2:

Perform the multiplication $50 \times 51$.

$$5 \left( \frac{2550}{2} \right)$$

Step 3.4.3:

Divide $2550$ by $2$.

$$5 \times 1275$$

Step 3.4.4:

Multiply $5$ by $1275$.

$$6375$$

Step 4:

Substitute the computed summations into the transformed expression.

$$25250 - 6375$$

Step 5:

Deduct $6375$ from $25250$ to get the final result.

$$18875$$

Knowledge Notes:

The problem involves evaluating an arithmetic series, which is a sequence of numbers with a constant difference between consecutive terms. The sum of an arithmetic series can be found using the formula:

$$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$$

This formula calculates the sum of the first $n$ natural numbers. When dealing with a summation that starts at a value other than $1$, we can split the summation into two parts: one that sums from $1$ to the upper limit, and another that sums from $1$ to one less than the starting value. The difference between these two sums gives us the sum from the original starting value to the upper limit.

In this problem, we also encounter the distributive property of multiplication over addition, which allows us to factor out constants from the summation. This simplifies the calculation by reducing the summation to a simpler form before applying the arithmetic series sum formula.

The steps taken in the solution involve algebraic manipulations such as factoring, substitution, and basic arithmetic operations (addition, subtraction, multiplication, and division). These are fundamental skills in algebra and are frequently used in solving summation and series problems.