1²+3²+5²....+ (2n-1)²=1/3n(4n²-1)solve?

To solve this, we can use the formula for the sum of the squares of the first n odd numbers:

1² + 3² + 5² + ... + (2n-1)² = 1/3n(4n² - 1)

First, let's simplify the left side of the equation:

1² + 3² + 5² + ... + (2n-1)² = (1 + 9 + 25 + ... + (2n-1)²)
= 1 + 9 + 25 + ... + (2n-1)²
= n² + (n-1)² + (n-2)² + ... + 1²

Now, let's substitute this into the formula:

n² + (n-1)² + (n-2)² + ... + 1² = 1/3n(4n² - 1)

Expanding the left side:

n² + (n-1)² + (n-2)² + ... + 1² = n² + (n-1)² + (n-2)² + ... + 1

Now, we can simplify the equation:

n² + (n-1)² + (n-2)² + ... + 1 = 1/3n(4n² - 1)

n(n+1)(2n+1)/6 = 1/3n(4n² - 1)

Now, we can solve for n:

n(n+1)(2n+1)/6 = 4n²/3 - 1/3

n(n+1)(2n+1) = 8n² - 2

2n³ + 3n² + n = 8n² - 2

2n³ - 5n² + n + 2 = 0

(n-2)(2n² - n - 1) = 0

n = 2 or n = (1 ± √9)/4

Therefore, the solutions for n are n = 2 or n = (1 ± 3)/4

So, the solutions for n are n = 2 or n = -1/4 or n = 1.25

Therefore, the solution for n is n = 2.

Step 1: Expand the Left-Hand Side (LHS)

LHS: 1² + 3² + 5² + ... + (2n-1)²

We can write this as:

LHS = Σ(k=1 to n) (2k-1)²

Step 2: Expand the Square

(2k-1)² = 4k² - 4k + 1

Step 3: Sum the Series

LHS = Σ(k=1 to n) (4k² - 4k + 1)

Step 4: Simplify the Sum

LHS = 4Σ(k=1 to n) k² - 4Σ(k=1 to n) k + Σ(k=1 to n) 1

The sum of k² is n(n+1)(2n+1)/6.

The sum of k is n(n+1)/2.

The sum of 1 is just n.

Therefore,

LHS = 4(n(n+1)(2n+1)/6) - 4(n(n+1)/2) + n

Step 5: Equate LHS and RHS

1/3n(4n²-1) = 4(n(n+1)(2n+1)/6) - 4(n(n+1)/2) + n

Step 6: Simplify and Solve

Multiplying both sides by 3n:

4n²-1 = 2n(n+1)(2n+1) - 6n(n+1) + 3n

Expanding and simplifying:

4n²-1 = 4n³ + 4n² + 2n³ + 2n² - 6n² - 6n + 3n

4n²-1 = 6n³ - 4n² + n

6n³ - 10n² + n + 1 = 0

Factoring:

(3n-1)(2n²+n+1) = 0

Since the quadratic term 2n² + n + 1 is always positive, 3n-1 = 0.

Therefore, n = 1/3.