Evaluate the Summation sum from n=10 to 19 of (n-3)^2
The question provides a mathematical expression that requires you to calculate the summation of a series. Specifically, it asks for the sum of the squares of the numbers from 10 to 19, each reduced by 3. In other words, you are to find the total sum when plugging each integer value n from 10 through 19 into the formula (n-3)^2, and adding up all those resulting values.
$\sum_{n = 10}^{19} \left(\left(\right. n - 3 \left.\right)\right)^{2}$
Answer
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:
Distributive Property (FOIL Method): This property allows us to expand expressions like $(a-b)^2$ into $a^2 - 2ab + b^2$.
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.
Combining Like Terms: This involves simplifying expressions by adding or subtracting coefficients of the same variable.
Breaking Down Complex Summations: We split the original summation into smaller parts that could be evaluated using the summation formulas.
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.
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.
