Answer :
To solve the problem of proving that:
[tex]\[1^3 + 3^3 + 5^3 + \cdots + (2n-1)^3 = n^2 \left(2n^2 - 1\right),\][/tex]
let's approach it step by step.
### Step 1: Understand the Sequence
The left-hand side (LHS) of the equation is the sum of the cubes of the first [tex]\(n\)[/tex] odd numbers:
[tex]\[ S = 1^3 + 3^3 + 5^3 + \cdots + (2n-1)^3. \][/tex]
### Step 2: Identify the General Terms
The [tex]\(k\)[/tex]-th odd number can be represented as [tex]\(2k - 1\)[/tex]. Therefore, the sequence is:
[tex]\[ (2 \times 1 - 1)^3, (2 \times 2 - 1)^3, (2 \times 3 - 1)^3, \ldots, (2 \times n - 1)^3. \][/tex]
### Step 3: Express the Sum
The sum can be expressed as:
[tex]\[ S = \sum_{k=1}^{n} (2k - 1)^3. \][/tex]
### Step 4: Expand the Cubes
Let's expand [tex]\((2k - 1)^3\)[/tex]:
[tex]\[ (2k - 1)^3 = 8k^3 - 12k^2 + 6k - 1. \][/tex]
### Step 5: Sum the Series
Summing these expanded terms from [tex]\(k = 1\)[/tex] to [tex]\(k = n\)[/tex]:
[tex]\[ S = \sum_{k=1}^{n} (8k^3 - 12k^2 + 6k - 1). \][/tex]
### Step 6: Separate the Summation
We can separate the sum into individual sums:
[tex]\[ S = 8 \sum_{k=1}^{n} k^3 - 12 \sum_{k=1}^{n} k^2 + 6 \sum_{k=1}^{n} k - \sum_{k=1}^{n} 1. \][/tex]
### Step 7: Use Summation Formulas
We use the known summation formulas:
[tex]\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2}, \][/tex]
[tex]\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}, \][/tex]
[tex]\[ \sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2, \][/tex]
[tex]\[ \sum_{k=1}^{n} 1 = n. \][/tex]
### Step 8: Substitute
Substitute these into the separated sum:
[tex]\[ S = 8 \left( \frac{n(n+1)}{2} \right)^2 - 12 \left( \frac{n(n+1)(2n+1)}{6} \right) + 6 \left( \frac{n(n+1)}{2} \right) - n. \][/tex]
### Step 9: Simplify Each Term
Simplify each term separately:
1. [tex]\( 8 \left( \frac{n(n+1)}{2} \right)^2 = 8 \frac{n^2(n+1)^2}{4} = 2n^2(n+1)^2. \)[/tex]
2. [tex]\( -12 \left( \frac{n(n+1)(2n+1)}{6} \right) = -2n(n+1)(2n+1). \)[/tex]
3. [tex]\( 6 \left( \frac{n(n+1)}{2} \right) = 3n(n+1). \)[/tex]
4. [tex]\( -n. \)[/tex]
### Step 10: Combine the Simplified Results
Combine and simplify:
[tex]\[ S = 2n^2(n^2 + 2n + 1) - 2n(n^2 + 3n + 2) + 3n(n + 1) - n. \][/tex]
### Step 11: Factor and Simplify Further
Combine similar terms and factor the expression:
[tex]\[ S = 2n^2n^2 + 4n^2n + 2n^2 - 2n^3 - 6n^2 - 4n + 3n^2 + 3n - n. \][/tex]
[tex]\[ S = 2n^4 - 2n^2. \][/tex]
Therefore,
[tex]\[ S = 2n^4 - n^2 = n^2(2n^2 - 1). \][/tex]
### Conclusion
Hence, we have proved that:
[tex]\[ 1^3 + 3^3 + 5^3 + \cdots + (2n-1)^3 = n^2 \left( 2n^2 - 1 \right). \][/tex]
[tex]\[1^3 + 3^3 + 5^3 + \cdots + (2n-1)^3 = n^2 \left(2n^2 - 1\right),\][/tex]
let's approach it step by step.
### Step 1: Understand the Sequence
The left-hand side (LHS) of the equation is the sum of the cubes of the first [tex]\(n\)[/tex] odd numbers:
[tex]\[ S = 1^3 + 3^3 + 5^3 + \cdots + (2n-1)^3. \][/tex]
### Step 2: Identify the General Terms
The [tex]\(k\)[/tex]-th odd number can be represented as [tex]\(2k - 1\)[/tex]. Therefore, the sequence is:
[tex]\[ (2 \times 1 - 1)^3, (2 \times 2 - 1)^3, (2 \times 3 - 1)^3, \ldots, (2 \times n - 1)^3. \][/tex]
### Step 3: Express the Sum
The sum can be expressed as:
[tex]\[ S = \sum_{k=1}^{n} (2k - 1)^3. \][/tex]
### Step 4: Expand the Cubes
Let's expand [tex]\((2k - 1)^3\)[/tex]:
[tex]\[ (2k - 1)^3 = 8k^3 - 12k^2 + 6k - 1. \][/tex]
### Step 5: Sum the Series
Summing these expanded terms from [tex]\(k = 1\)[/tex] to [tex]\(k = n\)[/tex]:
[tex]\[ S = \sum_{k=1}^{n} (8k^3 - 12k^2 + 6k - 1). \][/tex]
### Step 6: Separate the Summation
We can separate the sum into individual sums:
[tex]\[ S = 8 \sum_{k=1}^{n} k^3 - 12 \sum_{k=1}^{n} k^2 + 6 \sum_{k=1}^{n} k - \sum_{k=1}^{n} 1. \][/tex]
### Step 7: Use Summation Formulas
We use the known summation formulas:
[tex]\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2}, \][/tex]
[tex]\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}, \][/tex]
[tex]\[ \sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2, \][/tex]
[tex]\[ \sum_{k=1}^{n} 1 = n. \][/tex]
### Step 8: Substitute
Substitute these into the separated sum:
[tex]\[ S = 8 \left( \frac{n(n+1)}{2} \right)^2 - 12 \left( \frac{n(n+1)(2n+1)}{6} \right) + 6 \left( \frac{n(n+1)}{2} \right) - n. \][/tex]
### Step 9: Simplify Each Term
Simplify each term separately:
1. [tex]\( 8 \left( \frac{n(n+1)}{2} \right)^2 = 8 \frac{n^2(n+1)^2}{4} = 2n^2(n+1)^2. \)[/tex]
2. [tex]\( -12 \left( \frac{n(n+1)(2n+1)}{6} \right) = -2n(n+1)(2n+1). \)[/tex]
3. [tex]\( 6 \left( \frac{n(n+1)}{2} \right) = 3n(n+1). \)[/tex]
4. [tex]\( -n. \)[/tex]
### Step 10: Combine the Simplified Results
Combine and simplify:
[tex]\[ S = 2n^2(n^2 + 2n + 1) - 2n(n^2 + 3n + 2) + 3n(n + 1) - n. \][/tex]
### Step 11: Factor and Simplify Further
Combine similar terms and factor the expression:
[tex]\[ S = 2n^2n^2 + 4n^2n + 2n^2 - 2n^3 - 6n^2 - 4n + 3n^2 + 3n - n. \][/tex]
[tex]\[ S = 2n^4 - 2n^2. \][/tex]
Therefore,
[tex]\[ S = 2n^4 - n^2 = n^2(2n^2 - 1). \][/tex]
### Conclusion
Hence, we have proved that:
[tex]\[ 1^3 + 3^3 + 5^3 + \cdots + (2n-1)^3 = n^2 \left( 2n^2 - 1 \right). \][/tex]
