Evaluate the Summation sum from i=1 to 21 of i(i-1)^2
The problem is asking for the computation of a specified summation. Specifically, it requires the evaluation of a series where each term is the product of three factors: the index 'i', and the square of the index 'i' minus one. The summation involves calculating the value of this expression for each integer 'i' starting from 1 up to and including 21, and then adding up all these individual values to find the total sum.
$\sum_{i = 1}^{21} i \left(\left(\right. i - 1 \left.\right)\right)^{2}$
Answer
Solution:
Step 1: Simplify the given summation expression.
Step 1.1: Rewrite the squared term.
Rewrite $(i - 1)^2$ as $(i - 1)(i - 1)$ to get $i((i - 1)(i - 1))$.
Step 1.2: Expand the binomial using the FOIL method.
Step 1.2.1: Apply the distributive property.
Expand to get $i(i(i - 1) - (i - 1)(i - 1))$.
Step 1.2.2: Continue applying the distributive property.
Expand further to get $i(i^2 - i - (i - 1)(i - 1))$.
Step 1.2.3: Finalize the expansion.
Complete the expansion to get $i(i^2 - i - i + 1)$.
Step 1.3: Combine like terms.
Step 1.3.1: Simplify each term.
Step 1.3.1.1: Multiply $i$ by $i^2$.
This results in $i(i^2 - i - i + 1)$.
Step 1.3.1.2: Rearrange the terms.
Rearrange to get $i(i^2 - 2i + 1)$.
Step 1.3.1.3: Combine the $i$ terms.
Combine to get $i(i^2 - 2i + 1)$.
Step 1.4: Distribute the $i$ across the terms.
Multiply to get $i^3 - 2i^2 + i$.
Step 1.5: Simplify the expression.
Step 1.5.1: Apply exponent rules.
Step 1.5.1.1: Multiply $i$ by $i^2$.
Raise $i$ to the power of 3 to get $i^3$.
Step 1.5.1.2: Combine the exponents.
Combine to get $i^3$.
Step 1.5.2: Rewrite using commutative property.
Get $i^3 - 2i^2 + i$.
Step 1.5.3: Multiply $i$ by 1.
This results in $i^3 - 2i^2 + i$.
Step 1.6: Combine the $i$ terms.
Step 1.6.1: Combine the $i$ terms.
Get $i^3 - 2i^2 + i$.
Step 1.7: Rewrite the summation with the simplified expression.
Get $\sum_{i=1}^{21} (i^3 - 2i^2 + i)$.
Step 2: Break down the summation into separate summations.
$\sum_{i=1}^{21} i^3 - 2\sum_{i=1}^{21} i^2 + \sum_{i=1}^{21} i$
Step 3: Evaluate the cubic summation.
Step 3.1: Use the formula for the sum of cubes.
$\sum_{i=1}^{n} i^3 = \frac{n^2(n + 1)^2}{4}$
Step 3.2: Substitute the upper limit of the summation.
$\frac{21^2(21 + 1)^2}{4}$
Step 3.3: Simplify the expression.
Step 3.3.1: Simplify the numerator.
Multiply to get $\frac{441 \cdot 484}{4}$.
Step 3.3.2: Simplify the denominator.
Divide to get $53361$.
Step 4: Evaluate the squared summation.
Step 4.1: Use the formula for the sum of squares.
$\sum_{i=1}^{n} i^2 = \frac{n(n + 1)(2n + 1)}{6}$
Step 4.2: Substitute the upper limit and multiply by 2.
$-2 \cdot \frac{21(21 + 1)(2 \cdot 21 + 1)}{6}$
Step 4.3: Simplify the expression.
Step 4.3.1: Simplify the numerator.
Multiply to get $-2 \cdot \frac{462 \cdot 43}{6}$.
Step 4.3.2: Simplify the denominator.
Divide and multiply to get $-6622$.
Step 5: Evaluate the linear summation.
Step 5.1: Use the formula for the sum of first n natural numbers.
$\sum_{i=1}^{n} i = \frac{n(n + 1)}{2}$
Step 5.2: Substitute the upper limit.
$\frac{21(21 + 1)}{2}$
Step 5.3: Simplify the expression.
Step 5.3.1: Multiply and divide.
Get $231$.
Step 6: Combine the results of the summations.
$53361 - 6622 + 231$
Step 7: Simplify the final result.
Step 7.1: Subtract the second term.
$46739 + 231$
Step 7.2: Add the last term.
Final result is $46970$.
Knowledge Notes:
The solution involves several key knowledge points in algebra and summation:
Expansion of Binomials: The square of a binomial, $(a - b)^2$, is expanded using the FOIL (First, Outer, Inner, Last) method to get $a^2 - 2ab + b^2$.
Distributive Property: This property allows us to multiply a single term by each term inside a parenthesis, such as $a(b + c) = ab + ac$.
Combining Like Terms: Terms that have the same variable raised to the same power can be combined by adding or subtracting their coefficients.
Summation Formulas: There are formulas to find the sum of powers of natural numbers:
- Sum of first $n$ natural numbers: $\sum_{i=1}^{n} i = \frac{n(n + 1)}{2}$
- Sum of squares of first $n$ natural numbers: $\sum_{i=1}^{n} i^2 = \frac{n(n + 1)(2n + 1)}{6}$
- Sum of cubes of first $n$ natural numbers: $\sum_{i=1}^{n} i^3 = \frac{n^2(n + 1)^2}{4}$
Exponent Rules: When multiplying like bases, the exponents are added, such as $a^m \cdot a^n = a^{m+n}$.
Commutative Property of Multiplication: The order in which two numbers are multiplied does not affect the product, $ab = ba$.
Simplification: The process of reducing an expression to its simplest form by performing all possible operations and combining like terms.
