Evaluate the Summation sum from j=1 to 200 of 2j(j+3)

Solution:

Step 1: Simplify the given summation expression.

• Step 1.1: Distribute the multiplication across the terms: $2j \cdot j + 2j \cdot 3$

• Step 1.2: Perform the multiplication of $j$ with itself:

  • Step 1.2.1: Position the $j$ terms together: $2(j \cdot j) + 2j \cdot 3$
  • Step 1.2.2: Multiply $j$ by $j$: $2j^2 + 2j \cdot 3$

• Step 1.3: Multiply the constant $3$ by $2$: $2j^2 + 6j$

• Step 1.4: Represent the summation with the simplified terms: $\sum_{j=1}^{200} (2j^2 + 6j)$

Step 2: Break down the summation into two separate summations: $\sum_{j=1}^{200} (2j^2 + 6j) = 2\sum_{j=1}^{200} j^2 + 6\sum_{j=1}^{200} j$

Step 3: Calculate the summation of squared terms $2\sum_{j=1}^{200} j^2$.

• Step 3.1: Use the formula for the sum of squares: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

• Step 3.2: Plug in the values into the formula and include the coefficient $2$: $2\left(\frac{200(200+1)(2\cdot200+1)}{6}\right)$

• Step 3.3: Simplify the expression:

  • Step 3.3.1: Work on the numerator:

    • Step 3.3.1.1: Add $200$ and $1$: $2\frac{200\cdot201(2\cdot200+1)}{6}$

    • Step 3.3.1.2: Perform the multiplication steps:

      • Step 3.3.1.2.1: Multiply $200$ by $201$: $2\frac{40200(2\cdot200+1)}{6}$
      • Step 3.3.1.2.2: Multiply $2$ by $200$: $2\frac{40200(400+1)}{6}$

    • Step 3.3.1.3: Add $400$ and $1$: $2\frac{40200\cdot401}{6}$

  • Step 3.3.2: Reduce the fraction by cancelling common factors:

    • Step 3.3.2.1: Multiply $40200$ by $401$: $2\left(\frac{16120200}{6}\right)$

    • Step 3.3.2.2: Simplify by cancelling the $2$:

      • Step 3.3.2.2.1: Factor $2$ out of $6$: $2\frac{16120200}{2\cdot3}$
      • Step 3.3.2.2.2: Cancel out the $2$: $\frac{16120200}{3}$
      • Step 3.3.2.2.3: Rewrite the simplified expression: $\frac{16120200}{3}$

    • Step 3.3.2.3: Divide $16120200$ by $3$: $5373400$

Step 4: Calculate the summation of linear terms $6\sum_{j=1}^{200} j$.

• Step 4.1: Use the formula for the sum of the first $n$ natural numbers: $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$

• Step 4.2: Substitute the values into the formula and include the coefficient $6$: $6\left(\frac{200(200+1)}{2}\right)$

• Step 4.3: Simplify the expression:

  • Step 4.3.1: Simplify the terms:

    • Step 4.3.1.1: Add $200$ and $1$: $6\frac{200\cdot201}{2}$
    • Step 4.3.1.2: Multiply $200$ by $201$: $6\left(\frac{40200}{2}\right)$

  • Step 4.3.2: Cancel the common factor of $2$:

    • Step 4.3.2.1: Factor $2$ out of $6$: $3\cdot2\frac{40200}{2}$
    • Step 4.3.2.2: Cancel out the $2$: $3\cdot40200$
    • Step 4.3.2.3: Rewrite the expression: $3\cdot40200$

  • Step 4.3.3: Multiply $3$ by $40200$: $120600$

Step 5: Combine the results of the two summations: $5373400 + 120600$

Step 6: Add the two results together to find the final sum: $5494000$

Knowledge Notes:

• Summation Notation: Summation notation is a way to represent the addition of a sequence of numbers, typically expressed as $\sum$. The variable below the summation symbol indicates the starting index, and the number above indicates the ending index.

• Distributive Property: The distributive property states that $a(b + c) = ab + ac$. It allows us to multiply a number by a sum by multiplying each addend separately and then adding the products.

• Sum of Squares Formula: The sum of the squares of the first $n$ natural numbers is given by $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$. This formula is useful for finding the sum of squared terms in a series.

• Sum of Natural Numbers Formula: The sum of the first $n$ natural numbers is given by $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$. This formula is used to quickly calculate the sum of a sequence of consecutive integers.

• Simplification of Expressions: Simplification involves performing operations like addition, multiplication, and factoring to reduce an expression to its simplest form. This often includes cancelling common factors to reduce fractions.

• Combining Like Terms: When adding or subtracting algebraic expressions, we combine like terms, which are terms that have the same variables raised to the same power.

• Polynomial Summation: When dealing with the summation of polynomial expressions, it is often helpful to separate the summation into parts that correspond to each term of the polynomial, as long as each part follows the summation rules.