Evaluate the Summation sum from n=10 to 19 of (n-3)^2

Solution:

Step 1: Simplify the given summation expression.

• Step 1.1: Express $(n-3{)}^2$ as $(n-3)(n-3)$.

• Step 1.2: Expand the expression $(n-3)(n-3)$ using the distributive property (FOIL).

  • Step 1.2.1: Distribute $n$ across $(n-3)$ and $-3$ across $(n-3)$.

  • Step 1.2.2: Continue the distribution to get $n\cdot n-n\cdot 3-3\cdot n+3\cdot 3$.

  • Step 1.2.3: Finalize the distribution to obtain $n^2-3n-3n+9$.

• Step 1.3: Combine like terms in the expanded expression.

  • Step 1.3.1: Simplify each term individually.

    • Step 1.3.1.1: Calculate $n\cdot n$ to get $n^2$.

    • Step 1.3.1.2: Rearrange the terms to $n^2-3n-3n+9$.

    • Step 1.3.1.3: Multiply $-3$ by $-3$ to get $+9$.

  • Step 1.3.2: Combine the $-3n$ and $-3n$ to get $-6n$.

• Step 1.4: Rewrite the summation as $\sum _{n=10}^{19}(n^2-6n+9)$.

Step 2: Adjust the summation to start from $n=1$.

• $\sum _{n=10}^{19}(n^2-6n+9)=\sum _{n=1}^{19}(n^2-6n+9)-\sum _{n=1}^9(n^2-6n+9)$

Step 3: Evaluate the summation from $n=1$ to $n=19$.

• Step 3.1: Break down the summation into individual components.

  • $\sum _{n=1}^{19}(n^2-6n+9)=\sum _{n=1}^{19}{n}^2-6\sum _{n=1}^{19}n+\sum _{n=1}^{19}9$

• Step 3.2: Calculate $\sum _{n=1}^{19}{n}^2$.

  • Step 3.2.1: Use the formula $\sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}$.

  • Step 3.2.2: Plug in $n=19$ into the formula.

  • Step 3.2.3: Simplify the expression.

    • Step 3.2.3.1: Simplify the numerator.

      • Step 3.2.3.1.1: Calculate $19\cdot 20$.

      • Step 3.2.3.1.2: Multiply $2\cdot 19$ and add $1$.

    • Step 3.2.3.2: Simplify the denominator.

      • Step 3.2.3.2.1: Multiply $380\cdot 39$.

      • Step 3.2.3.2.2: Divide $14820$ by $6$ to get $2470$.

• Step 3.3: Calculate $6\sum _{n=1}^{19}n$.

  • Step 3.3.1: Use the formula $\sum _{k=1}^{n}k=\frac{n(n+1)}{2}$.

  • Step 3.3.2: Multiply the result by $-6$ after substituting $n=19$.

  • Step 3.3.3: Simplify to get $-1140$.

• Step 3.4: Calculate $\sum _{n=1}^{19}9$.

  • Step 3.4.1: Use the formula $\sum _{k=1}^{n}c=cn$.

  • Step 3.4.2: Multiply $9$ by $19$ to get $171$.

• Step 3.5: Combine the results of the individual summations.

• Step 3.6: Simplify to get the final result of $1501$.

Step 4: Evaluate the summation from $n=1$ to $n=9$.

• Step 4.1: Break down the summation into individual components.

  • $\sum _{n=1}^9(n^2-6n+9)=\sum _{n=1}^9{n}^2-6\sum _{n=1}^{9}n+\sum _{n=1}^{9}9$

• Step 4.2: Calculate $\sum _{n=1}^9{n}^2$.

  • Step 4.2.1: Use the formula $\sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}$.

  • Step 4.2.2: Plug in $n=9$ into the formula.

  • Step 4.2.3: Simplify to get $285$.

• Step 4.3: Calculate $6\sum _{n=1}^{9}n$.

  • Step 4.3.1: Use the formula $\sum _{k=1}^{n}k=\frac{n(n+1)}{2}$.

  • Step 4.3.2: Multiply the result by $-6$ after substituting $n=9$.

  • Step 4.3.3: Simplify to get $-270$.

• Step 4.4: Calculate $\sum _{n=1}^{9}9$.

  • Step 4.4.1: Use the formula $\sum _{k=1}^{n}c=cn$.

  • Step 4.4.2: Multiply $9$ by $9$ to get $81$.

• Step 4.5: Combine the results of the individual summations.

• Step 4.6: Simplify to get the final result of $96$.

Step 5: Combine the results from Steps 3 and 4.

• Replace the summations with the values found: $1501-96$.

Step 6: Subtract to find the final answer.

• Subtract $96$ from $1501$ to get $1405$.

Knowledge Notes:

To solve the given problem, we used several mathematical concepts and formulas:

1. Distributive Property (FOIL Method): This property allows us to expand expressions like $(a-b{)}^2$ into $a^2-2ab+b^2$.

2. Summation Formulas: We used standard summation formulas for squares of natural numbers and arithmetic series:

   • $\sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}$ for the sum of squares.

   • $\sum _{k=1}^{n}k=\frac{n(n+1)}{2}$ for the sum of the first $n$ natural numbers.

   • $\sum _{k=1}^{n}c=cn$ for the sum of a constant $c$ repeated $n$ times.

3. Combining Like Terms: This involves simplifying expressions by adding or subtracting coefficients of the same variable.

4. Breaking Down Complex Summations: We split the original summation into smaller parts that could be evaluated using the summation formulas.

5. Adjusting Summation Limits: We adjusted the summation limits to start from 1 to utilize the summation formulas, then subtracted the unwanted part of the summation.

6. Simplification of Fractions: This involved reducing fractions to their simplest form by canceling common factors.

By applying these concepts, we were able to evaluate the summation of $(n-3{)}^2$ from $n=10$ to $19$ and find the final result.